text:
\section{Definitions}
Let $q$ be a complex number with $0<|q|<1$ and define $\mathbb{C}^*:=\mathbb{C}-\{0\}$.  We have
{\allowdisplaybreaks \begin{gather}
(x)_n=(x;q)_n:=\prod_{i=0}^{n-1}(1-q^ix), \ \ (x)_{\infty}=(x;q)_{\infty}:=\prod_{i\ge 0}(1-q^ix),\notag\\
 j(x;q):=(x)_{\infty}(q/x)_{\infty}(q)_{\infty}=\sum_{n=-\infty}^{\infty}(-1)^nq^{\binom{n}{2}}x^n,\label{equation:theta-series}\\
 {\text{and }}\ \ j(x_1,x_2,\dots,x_n;q):=j(x_1;q)j(x_2;q)\cdots j(x_n;q).\notag
\end{gather}}%
where in the middle line the equivalence of product and sum follows from Jacobi's triple product identity.  Identity (\ref{equation:theta-series}) defines a theta function.  A {\em partial} theta function is half of a theta function, for example
\begin{equation}
\sum_{n=0}^{\infty}(-1)^nq^{\binom{n}{2}}x^n.\label{equation:partial-theta}
\end{equation}
Let $a$ and $m$ be integers with $m$ positive.  We further define
\begin{gather}
J_{a,m}:=j(q^a;q^m), \ \ \overline{J}_{a,m}:=j(-q^a;q^m), \ {\text{and }}J_m:=J_{m,3m}=\prod_{i\ge 1}(1-q^{mi}).
\end{gather}

\section{Introduction}\label{section:introduction}
Freeman Dyson conjectured a beautiful combinatorial description of Ramanujan's congruences for the partition function using a statistic which he called the rank.  The rank has further intrigue due to its connections with Ramanujan's mock theta functions.   Atkin and Swinnerton-Dyer proved Dyson's conjectures and gave additional examples of how the rank relates to theta and mock theta functions.  Later, Bringmann and Ono generalised Atkin and Swinnerton-Dyer's results by using the theory of harmonic weak Maass forms, a theory which generalises mock theta functions.  In this paper, we use Appell--Lerch sum properties to give an explicit formula for Dyson's ranks which then yields as straightforward consequence results of Atkin and Swinnerton-Dyer \cite{ASD} as well as Bringmann, Ono, Rhoades, and Zagier \cite{BrO2, BrOR, Za}.

In Ramanujan's last letter to Hardy, he gave a list of seventeen functions which he called ``mock theta functions''  \cite{R1}.   Each mock theta function $f(q)$ was defined as a $q$-series, convergent  for $|q|<1$, such that for every root of unity $\zeta$, there is a theta function $\theta_{\zeta}(q)$ such that the difference $f(q)-\theta_{\zeta}(q)$ is bounded as $q\rightarrow \zeta$ radially; moreover, there is no single theta function which works for all $\zeta$ \cite{GOR}.   Ramanujan also stated identities relating mock theta functions to each other as well as to modular forms, which suggested that mock theta functions live inside of a vector space, which has a subspace consisting of modular forms.  Later, more mock theta identities were found in the Lost Notebook \cite{RLN, H1}

The Lost Notebook is the Rosetta Stone of $q$-series.  Indeed, numerous entries expand $q$-hypergeometric series in terms of theta functions (Rogers-Ramanujan type identities), Appell--Lerch sums (mock theta functions), or partial theta functions.   Partial theta functions  play roles in areas outside of number theory such as quantum invariants of $3$-manifolds \cite{LZ}.  Appell--Lerch sums appear in the context of black hole physics \cite{DMZ}.  Long-standing problems have been to determine the modularity of the mock thetas and to understand how various types of $q$-series representations relate to each other.

In Zwegers' breakthrough work \cite{Zw2}, he solved the modularity question for mock theta functions.    Although mock theta functions are not modular \cite{W3}, he showed that they can be completed to non-holomorphic functions which are modular.  As a result, mock theta functions may be viewed as holomorphic parts of harmonic weak Maass forms \cite{  BrO1, BrO2,BruF, Za}.

\subsection{Dyson's ranks}

One way to study mock theta functions is through partitions.  A partition of a positive integer $n$ is a weakly-decreasing sequence of positive integers whose sum is $n$.    The partitions of $4$ are $(4),$ $(3,1),$ $(2,2),$ $(2,1,1),$ $(1,1,1,1)$.    We denote the number of partitions of $n$ by $p(n)$.  Among the most famous results in partitions are Ramanujan's congruences:
{\allowdisplaybreaks \begin{align*}
p(5n+4)\equiv &\  0 \pmod{5}, \\
p(7n+5)\equiv &\  0 \pmod{7} ,\\
p(11n+6)\equiv &\  0 \pmod{11},
\end{align*}}%
\noindent for which there are many proofs and generalisations.  To study partition congruences, one often constructs a function, called a statistic, which assigns an integer value to a partition.  Dyson \cite{D1} gave insight into the first two congruences with such a statistic, which he called the rank.  He defined the rank of a partition to be the largest part minus the number of parts.   In particular, the ranks of the five partitions of $4$ are $3,1,0,-1,-3,$ respectively, giving an equinumerous distribution of the partitions of $4$ into the five distinct residue classes mod $5$.  In general, one defines
\begin{equation*}
N(a,M;n) := \mathrm{number\ of\ partitions\ of\ } n \mathrm{\ with\ rank\ } \equiv a \ (\mathrm{mod\ } M),
\end{equation*}
which has the symmetry property $N(a,M;n)=N(M-a,M;n)$.  Dyson conjectured \cite{D1}, \cite[$(2.2)$--$(2.11)$]{ASD}:
\begin{subequations}
\begin{equation}
N(1,5;5n+1)=N(2,5;5n+1)\label{equation:ASD-2.2},
\end{equation}
\begin{equation}
N(0,5;5n+2)=N(2,5;5n+2),
\end{equation}
\begin{equation}
N(0,5;5n+4)=N(1,5;5n+4)=N(2,5;5n+4),\label{equation:Dyson-M5}
\end{equation}
\begin{equation}
N(2,7;7n)=N(3,7;7n),\label{equation:ASD-2.5}
\end{equation}
\begin{equation}
N(1,7;7n+1)=N(2,7;7n+1)=N(3,7,7n+1),\label{equation:ASD-2.6}
\end{equation}
\begin{equation}
N(0,7,7n+2)=N(3,7;7n+2),
\end{equation}
\begin{equation}
N(0,7;7n+3)=N(2,7;7n+3), \ \ N(1,7;7n+3)=N(3,7;7n+3),
\end{equation}
\begin{equation}
N(0,7;7n+4)=N(1,7;7n+4)=N(3,7;7n+4),
\end{equation}
\begin{equation}
N(0,7;7n+5)=N(1,7;7n+5)=N(2,7;7n+5)=N(3,7;7n+5),\label{equation:Dyson-M7}
\end{equation}
\begin{equation}
N(0,7;7n+6)+N(1,7;7n+6)=N(2,7;7n+6)+N(3,7;7n+6).\label{equation:ASD-2.11}
\end{equation}
\end{subequations}
In particular, indentities (\ref{equation:Dyson-M5}) and (\ref{equation:Dyson-M7}) with the symmetry property give the first two of Ramanujan's congruences.

We define the general rank-difference
\begin{equation}
R(a,b,M,c,m):=\sum_{n= 0}^{\infty} \Big ( N(a,M;mn+c)-N(b,M;mn+c)\Big )q^n,\label{equation:R-def}
\end{equation}
where $a,b,c,m,M$ are integers with $0\le a,b<M$ and $0\le c<m$.  Dyson's conjectures (\ref{equation:ASD-2.2})--(\ref{equation:Dyson-M7}) can be written as rank-differences that are equal to zero.  Atkin and Swinnerton-Dyer not only proved Dyson's conjectures but also determined rank-differences \cite[Theorems $4$, $5$]{ASD} which are equal to theta and mock theta functions.  As an example of Atkin and Swinnerton-Dyer's other results, they proved for modulus $5$ \cite[Theorem $4$]{ASD}, slightly rewritten,
\begin{subequations}
\begin{equation}
R(1,2,5,0,5)=qg(q,q^5),\label{equation:ASD-6.11}
\end{equation}
\begin{equation}
R(0,2,5,0,5)+2R(1,2,5,0,5)={J_{1}J_{5}^3}/{J_{1,5}^3},\label{equation:ASD-6.12}
\end{equation}
\begin{equation}
R(0,2,5,1,5)={J_{5}^2}/{J_{1,5}},\label{equation:ASD-6.13}
\end{equation}
\begin{equation}
R(1,2,5,2,5)=J_{5}^2/J_{2,5},\label{equation:ASD-6.16}
\end{equation}
\begin{equation}
R(0,2,5,3,5)=-qg(q^2,q^5),\label{equation:ASD-6.17}
\end{equation}
\begin{equation}
R(0,1,5,3,5)+R(0,2,5,3,5)=J_{1}J_{5}^3/J_{2,5}^3,\label{equation:ASD-6.18}
\end{equation}
\end{subequations}
where
\begin{equation}
g(x,q):=x^{-1}\Big ( -1+\sum_{n= 0}^{\infty}\frac{q^{n^2}}{(x;q)_{n+1}(q/x;q)_n}\Big )\label{equation:g-def}
\end{equation}
is a universal mock theta function.

Many of Ramanujan's mock theta functions can be written in terms of rank-differences.  For two fifth order mock theta functions \cite{H1}:
{\allowdisplaybreaks\begin{subequations}
\begin{equation}
\chi_0(q):=\sum_{n= 0}^{\infty}\frac{q^n}{(q^{n+1})_n}=2+3qg(q,q^5)-\frac{J_5^2J_{2,5}}{J_{1,5}^2}=2+R(1,0,5,0,5),\label{equation:chi0}
\end{equation}
\begin{equation}
\chi_1(q):=\sum_{n= 0}^{\infty}\frac{q^n}{(q^{n+1})_{n+1}}=3qg(q^2,q^5)+\frac{J_5^2J_{1,5}}{J_{2,5}^2}=R(2,0,5,3,5)+R(2,1,5,3,5).\label{equation:chi1}
\end{equation}
\end{subequations}}%
Most results about rank-differences have $m=M$, see (\ref{equation:R-def}), but for the mock theta $f(q)$:
\begin{equation}
f(q):=\sum_{n= 0}^{\infty}\frac{q^{n^2}}{(-q)_n^2}=2-2g(-1,q)=R(0,1,2,0,1).\label{equation:3rd-f}
\end{equation}
So we see that the $n$-th Fourier coefficient of $f(q)$ is equal to the number of partitions of $n$ with even rank minus the number of partitions of $n$ with odd rank \cite{BrO1}.

\subsection{The Main Theorem:  a special case}  We give an example of how our Main Theorem \ref{theorem:Theorem} yields Dyson's conjectures and Atkin and Swinnerton-Dyer's results.  For integers $0\le a<M$, define
\begin{equation}
D(a,M) = D(a, M,q) := \sum_{n= 0}^{\infty}\left(N(a,M;n) - \frac{p(n)}{M}\right) q^n.\label{equation:D-def}
\end{equation}
In terms of our definition (\ref{equation:R-def})
\begin{equation}
R(a,b,M,0,1)=D(a,M)-D(b,M). \label{equation:R2D}
\end{equation}
The Main Theorem \ref{theorem:Theorem} decomposes (\ref{equation:D-def}) into modular and mock modular components.  For modulus $M=5$, the Main Theorem specialises to
{\allowdisplaybreaks \begin{align}
D(0,5)&=-2 q^5 g(q^5, q^{25})\label{equation:D05}\\
& \ \ \ \ + \frac{4}{5}\cdot \frac{J_5 J_{25}^3}{J_{5,25}^3}
+ \frac{4}{5}\cdot q^1\cdot \frac{J_{25}^2}{J_{5,25}}
- \frac{2}{5}\cdot q^2\cdot \frac{ J_{25}^2}{J_{10,25}}
+ \frac{2}{5}\cdot q^3\cdot \frac{J_{5} J_{25}^3}{J_{10,25}^3}\notag\\
D(1,5) &= D(4,5) = q^5 g(q^5, q^{25})  - q^8 g(q^{10}, q^{25})\label{equation:D15-D45}\\
& \ \ \ \  - \frac{1}{5} \cdot    \frac{J_{5} J_{25}^3}{J_{5,25}^3}
  - \frac{1}{5} \cdot q^1 \cdot \frac{  J_{25}^2}{J_{5,25}}
 + \frac{3}{5} \cdot q^2 \cdot \frac{ J_{25}^2}{J_{10,25}}
 - \frac{3}{5} \cdot q^3 \cdot \frac{ J_5 J_{25}^3}{J_{10,25}^3}\notag\\
D(2,5) &= D(3,5) = q^8 g(q^{10}, q^{25}) \label{equation:D25-D35}\\
& \ \ \ \  - \frac{1}{5}\cdot \frac{J_5 J_{25}^3}{J_{5,25}^3}
- \frac{1}{5}\cdot q^1 \cdot \frac{ J_{25}^2}{J_{5,25}}
- \frac{2}{5}\cdot q^2 \cdot \frac{ J_{25}^2}{J_{10,25}}
+ \frac{2}{5}\cdot q^3 \cdot \frac{ J_5 J_{25}^3}{J_{10,25}^3}.\notag
\end{align}}%
The general form of the Main Theorem has twelve cases depending on $a$ and $M$ modulo $3$.

Identities (\ref{equation:ASD-2.2})--(\ref{equation:Dyson-M5}) and (\ref{equation:ASD-6.11})--(\ref{equation:ASD-6.18}) are now immediate.  Because of definitions (\ref{equation:g-def}) and (\ref{equation:theta-series}), the terms in the Fourier expansions from the $g$'s and $J$'s will be in terms of powers of $q^5$.  Given that there are no $q^{5k+4}$ terms in the expansions  (\ref{equation:D05}) -- (\ref{equation:D25-D35}), it is easy to see the first of Ramanujan's congruences (\ref{equation:Dyson-M5}).  We also see that (\ref{equation:ASD-2.2}) follows from (\ref{equation:D15-D45}) and (\ref{equation:D25-D35}) since the  coefficients of the $g$ parts do not involve any powers of $q$ of the form $q^{5k+1}$ and the theta parts for those powers are both equal to $-1/5 \cdot q^1 \cdot J_{25}^2/J_{5,25}$.   We even see rank-differences (\ref{equation:chi0}) and (\ref{equation:chi1}).

To prove and use our Main Theorem, we will need a building block finer than the universal mock theta function $g(x,q)$.  Although all of the mock theta functions from Ramanujan's letter \cite{R1} can be expressed in terms of the universal mock theta function $g(x,q)$ \cite{W3, H1,H2}, many mock thetas found in the Lost Notebook \cite{RLN} cannot.  In \cite{HM}, we demonstrated that all of the classical mock theta functions \cite{R1, RLN} can be expressed in terms of Appell-Lerch sums \cite{L1} defined as follows.  Let $x$, $q$, and $z$ be nonzero complex numbers with $|q|<1$ and neither $z$ nor $zx$ equal to an integral power of $q$, then
\begin{equation}
m(x,q,z):=\frac{1}{j(z;q)}\sum_{r=-\infty}^{\infty}\frac{(-1)^rq^{\binom{r}{2}}z^r}{1-q^{r-1}xz}.\label{equation:mxqz-def}
\end{equation}
From \cite[Theorem $2.2$]{H1} and \cite{HM}, we can express $g(x,q)$ in terms of $m(x,q,z):$
\begin{align}
g(x,q)&=-x^{-2}m(qx^{-3},q^3,x^3z)-x^{-1}m(q^2x^{-3},q^3,x^3z)+\frac{J_1^2j(xz;q)j(z;q^3)}{j(x;q)j(z;q)j(x^3z;q^3)},\label{equation:g-full}
\end{align}
where the right-hand side is actually $z$-independent.

The Main Theorem can be used to determine the arithmetic progressions $\mathcal{A}$ for which
\begin{equation}
\sum_{n\in \mathcal{A}, \ n\ge 0} \Big (N(a,M;n)-\frac{p(n)}{M}\Big )q^{n-{1}/{24}}\label{equation:Zag-def}
\end{equation}
is modular.  To sketch an example, take expansions (\ref{equation:D05}) -- (\ref{equation:D25-D35}).  The mock modular contributions are supported on arithmetic progressions of modulus $5$, e.g. for (\ref{equation:D05}) the mock contribution is supported on the $q$-terms of the form $q^{5n}$.  As a consequence, an arithmetic progression modulo $5$ from (\ref{equation:D05}) not containing the terms $q^{5n}$ will give a weakly holomorphic modular form.  Identity (\ref{equation:g-full}) and an Appell--Lerch sum property \cite[Theorems $3.5$]{HM}, which we recall in Section \ref{section:tech}, allows us to control the modulus of the progression on which the mock modular component is supported.  A detailed discussion is found in Section \ref{section:tech}.

\subsection{Harmonic weak Maass forms}

It turns out that our Main Theorem gives as a straightforward consequence recent celebrated results of Bringmann, Ono, and Rhoades \cite{BrO2, BrOR, Za}, in which they employ the theory of harmonic weak Maass forms in order to produce far-reaching generalizations of Dyson's and Atkin and Swinnerton-Dyer's observations on rank-differences. To give examples of their results we recall notation.  For an integer $t>0$, define $\mathfrak{f}_t:=\frac{2t}{\gcd (t,6)}$, $\mathfrak{l}_t:={\textup{lcm}}(2t^2,24)$, and $\tilde{\mathfrak{l}}_t:=\mathfrak{l}_t/24.$  Define the group $\Gamma_c$ by
\begin{equation}
\Gamma_c:=\Big < \Big ( \begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}\Big ),  \Big ( \begin{matrix} 1 & 0 \\ \mathfrak{l}_c^2 & 0 \end{matrix}\Big )\Big >.
\end{equation}
Bringmann and Ono showed that the deviation of the ranks from the average value is a mock theta function:

\begin{theorem}  {\cite[Theorem $1.3$]{BrO2}} If $0\le r <t$ are integers, then
\begin{equation}
\sum_{n=0}^{\infty}\Big (N(r,t;n)-\frac{p(n)}{t}\Big )q^{\mathfrak{l}_t n-\tfrac{\mathfrak{l}_t}{24}}\label{equation:BrO2-Thm1.3-eqn}
\end{equation}
is a holomorphic part of a weak Maass form of weight $1/2$ on  $\Gamma_t$.  Moreover, if $t$ is odd, then it is on $\Gamma_{1}\big ( 144 \ \mathfrak{f}_t^2 \ \tilde{\mathfrak{l}}_t\big ).$
\end{theorem}
\noindent They found arithmetic progressions of the deviation that are modular:
\begin{theorem}  {\cite[Theorem $1.4$]{BrO2}} If $0\le r <t$ are integers, where $t$ is odd, and $\mathcal{P}\nmid 6t$ is prime, then
\begin{equation}
\sum_{\substack{n\ge 1 \\ \big ( \frac{24\mathfrak{l}_t n -\mathfrak{l}_t}{\mathcal{P}}\big )=-\big( \frac{-24\tilde{\mathfrak{l}}_t}{\mathcal{P}}\big )}}\Big (N(r,t;n)-\frac{p(n)}{t}\Big )q^{\mathfrak{l}_t n-\tfrac{\mathfrak{l}_t}{24}}\label{equation:BrO2-Thm1.4-eqn}
\end{equation}
is a weight $1/2$ weakly holomorphic modular form on $\Gamma_{1}\big ( 144 \ \mathfrak{f}_t^2 \  \tilde{\mathfrak{l}}_t \ \mathcal{P}^4\big ).$
\end{theorem}
\begin{remark}
One could rewrite (\ref{equation:BrO2-Thm1.4-eqn}) as a sum over $n$ with $ \big ( \frac{1-24n}{\mathcal{P}}\big )=-1$.
\end{remark}

\noindent Bringmann, Ono,  and Rhoades, also found generalisations of Atkin and Swinnerton-Dyer's results on rank-differences, see also Zagier \cite{Za}:
\begin{theorem} {\cite[Theorem $1.1$]{BrOR}} Suppose that $t\ge 5$ is prime, $0\le r_1,r_2 < t$ and $0\le d < t$.  Then the following are true:
\begin{itemize}
\item[(1)] If $\big ( \frac{1-24d}{t} \big ) =-1$, then
\begin{equation*}
\sum_{n=0}^{\infty} (N(r_1,t;tn+d)-N(r_2,t;tn+d))q^{24(tn+d)-1}
\end{equation*}
is a weight $1/2$ weakly holomorphic modular form on $\Gamma_1(576t^6).$
\item[(2)] Suppose that $\big ( \frac{1-24d}{t} \big ) =1$.  If $r_1,r_2\not \equiv \tfrac{1}{2}(\pm 1 \pm \alpha) \pmod t$, where $\alpha$ is any integer for which $0\le \alpha <2t$ and $1-24d\equiv \alpha ^2 \pmod {2t}$, then
\begin{equation*}
\sum_{n=0}^{\infty} (N(r_1,t;tn+d)-N(r_2,t;tn+d))q^{24(tn+d)-1}
\end{equation*}
is a weight $1/2$ weakly holomorphic modular form on $\Gamma_1(576t^6).$
\end{itemize}
\end{theorem}

To obtain their results,  Bringmann et al. first complete an expression similar to (\ref{equation:Zag-def}) to a weight $1/2$ harmonic weak Maass form.  They then use a generalisation of quadratic twists (see for example Proposition $22$ of \cite{Ko}) in order to obtain an expression in which the non-holomorphic terms are supported on a certain arithmetic progression.  Using quadratic twists again they eliminate the non-holomorphic terms to obtain that, say, (\ref{equation:BrO2-Thm1.4-eqn}) is a weakly holomorphic modular form.  Using orthogonality of Dirichlet characters they then refine their results to, for example, \cite[Theorem $1.1(1)$]{BrOR}.

\subsection{Results}

We obtain more explicit results without appealing to the theory of harmonic weak Maass forms.  Where Bringmann et al first complete (\ref{equation:Zag-def}) to a weight $1/2$ harmonic weak Maass form, we use Appell--Lerch sum properties to decompose (\ref{equation:Zag-def}) into modular and mock modular components, similar to (\ref{equation:D05}) -- (\ref{equation:D25-D35}).   Thus the Main Theorem is stronger version of \cite[Theorem $1.3$]{BrO2}.  To construct an arithmetic progression of terms from (\ref{equation:Zag-def}) which is a weakly holomorphic modular form, one applies quadratic twists and orthogonality of Dirichlet characters to the modular component.  One then checks that the mock modular component of the decomposition is supported on a disjoint progression.

In Section \ref{section:tech} we discuss the role Appell--Lerch sums and their properties \cite[Theorems $3.9$, $3.5$]{HM} play in the proof the Main Theorem and in its applications.  In Section \ref{section:theorem} we state the Main Theorem and give additional specializations relevant to Dyson's rank-differences \cite{ASD, D1}.  In Section \ref{section:proof} we prove the Main Theorem using classical methods.  In Section \ref{section:BrO} we demonstrate how the Main Theorem yields results of Bringmann et al \cite{BrO2, BrOR, Za} on Dyson's ranks and Maass forms.   In Section \ref{section:M2} we apply our techniques to the $M_2$ rank of Berkovitch and Garvan \cite{BG}, which is based on work of Dyson \cite{D2,D3}.

\section{The Role of Appell--Lerch sums}\label{section:tech}

We first recall well-known Appell--Lerch sum properties.
 \begin{proposition}  \cite{L1, HM, Zw2} If $x$, $q$, $z$, $z_0$, and $z_1$ satisfy the appropriate conditions of definition (\ref{equation:mxqz-def}), then
{\allowdisplaybreaks \begin{subequations}
\begin{equation}
m(x,q,z)=m(x,q,qz),\label{equation:m-fnq-z}
\end{equation}
\begin{equation}
m(x,q,z)=x^{-1}m(x^{-1},q,z^{-1}),\label{equation:m-fnq-flip}
\end{equation}
\begin{equation}
m(qx,q,z)=1-xm(x,q,z),\label{equation:m-fnq-x}
\end{equation}
\begin{equation}
m(x,q,z_1)-m(x,q,z_0)=\frac{z_0J_1^3j(z_1/z_0;q)j(xz_0z_1;q)}{j(z_0;q)j(z_1;q)j(xz_0;q)j(xz_1;q)}.\label{equation:m-change-z}
\end{equation}
\end{subequations}}
\end{proposition}
In our work \cite{HM}, we introduced a heuristic relating partial theta functions to Appell--Lerch sums.  The heuristic suggests that identities involving partial theta functions have analogous identities in terms of Appell--Lerchs sums; one just needs to add a theta function \cite[Section $3$]{HM}, \cite{Mo1}.  So for example, trivial identities such as
{\allowdisplaybreaks \begin{gather}
\sum_{n=0}^{\infty}(-1)^nx^nq^{\binom{n}{2}}-\sum_{n=0}^{\infty}(-1)^n(-x)^nq^{\binom{n}{2}}
=-2x\sum_{n=0}^{\infty}(-1)^n(-x^2q^3)^{n}q^{4\binom{n}{2}},\label{equation:analog-1}\\
\sum_{n=0}^{\infty}(-1)^nx^nq^{\binom{n}{2}}=\sum_{n=0}^{\infty}(-1)^n(-x^2q)^{n}q^{4\binom{n}{2}}-x\sum_{n=0}^{\infty}(-1)^n(-x^2q^3)^{n}q^{4\binom{n}{2}},\label{equation:analog-2}
\end{gather}}%
where we sum over roots of unity or break up the summation index modulo some number, e.g. $2$, have non-trivial analogs in terms of Appell--Lerch sums.  The analog of (\ref{equation:analog-1}) generalises to

\begin{theorem} \cite[Theorem $3.9$] {HM} \label{theorem:rootsof1}
Let $n$ and $k$ be integers with $0 \leq k < n$. Let $\omega$ be a primitive $n$-th root of unity. Then
\begin{align*}
\sum_{t=0}^{n-1}& \omega^{-kt} m(\omega^t x,q,z) =  n q^{-\binom{k+1}{2}} (-x)^k m\big(-q^{{\binom{n}{2}-nk}} (-x)^n, q^{n^2}, z' \big)+n\Psi_k^n(x,z,z^{\prime};q)
\end{align*}
where
\begin{align}
\Psi_k^n&(x,z,z^{\prime};q):=\label{equation:msplit-theta}\\
&-\frac{ x^k z^{k+1} J_{n^2}^3} {j(z;q)j(z';q^{n^2})}\cdot\sum_{t=0}^{n-1} \frac{q^{{\binom{t+1}{2}+kt}} (-z)^t j\big(-q^{{\binom{n+1}{2}+nk+nt}} (-z)^n/z';q^{n^2}\big)
j(q^{nt} x^n z^n z';q^{n^2})}
{j\big(-q^{{\binom{n}{2}-nk}} (-x)^n z', q^{nt} x^n z^n;q^{n^2}\big)}.\notag
\end{align}
\end{theorem}

Let $\zeta_M:=e^{2\pi i /M}$, we have from \cite[$(3.13)$]{BrO2}:
\begin{align}
D(a,M)=\frac{1}{M}\sum_{j=1}^{M-1}\zeta_M^{-aj}\Big (1-\zeta_M^j\Big )\Big (1+\zeta_M^jg(\zeta_M^j,q)\Big ),\label{equation:D-to-g}
\end{align}
where the $g(x,q)$ is the universal mock theta function (\ref{equation:g-def}).  To decompose $D(a,M)$ into modular and mock modular components, we first expand (\ref{equation:D-to-g}) in terms of the $m(x,q,z)$ function, see (\ref{equation:g-full}).  Once expanded, we can use  Thereom \ref{theorem:rootsof1} to collapse the sum over roots of unity.  We then use (\ref{equation:g-full}) again to write the resulting $m(x,q,z)$ functions  in terms of the universal mock theta function $g(x,q)$.   This method gives our Main Theorem.

Producing generalisations of rank-differences such as  \cite[Theorem $1.4$]{BrO2} and \cite[Theorem $1.1$]{BrOR} requires another Appell--Lerch sum property from \cite{HM}.  Let us consider (\ref{equation:D25-D35}).  We remind the reader of the various ways of producing new modular forms from old ones such as by taking the twist by a Dirichlet character or by restricting the index of summation of the Fourier expansion to terms that lie in certain arithmetic progressions.  With this in mind we see that up to multiplication by a factor $q^{-1/24}$ that we can restrict the right-hand side (\ref{equation:D25-D35}) to the sequences $q^{5n}$, $q^{5n+1}$, or $q^{5n+2}$, and obtain a weakly holomorphic modular form.  In particular, we point out that these are the sequences to which $q^{8}g(q^{10},q^{25})$ does not contribute.  If we wish to consider sequences modulo $M$ instead of modulo $5$,  we rewrite the $q^{8}g(q^{10},q^{25})$ in terms of the $m(x,q,z)$ function, see (\ref{equation:g-full}), and use the following theorem with $n=M$ and $z^{\prime}=-1$:
\begin{theorem} \cite[Theorem $3.5$]{HM} \label{theorem:msplit-general-n} For generic $x,z,z'\in \mathbb{C}^*$
{\allowdisplaybreaks \begin{align*}
m(&x,q,z) = \sum_{r=0}^{n-1} q^{{-\binom{r+1}{2}}} (-x)^r m\big(-q^{{\binom{n}{2}-nr}} (-x)^n, q^{n^2}, z' \big)\\\notag
&+ \frac{z' J_n^3}{j(xz;q) j(z';q^{n^2})} \cdot \sum_{r=0}^{n-1}
\frac{q^{{\binom{r}{2}}} (-xz)^r
j\big(-q^{{\binom{n}{2}+r}} (-x)^n z z';q^n\big)
j(q^{nr} z^n/z';q^{n^2})}
{ j\big(-q^{{\binom{n}{2}}} (-x)^n z', q^r z;q^n\big )}.
\end{align*}}%
\end{theorem}
\noindent Instead of only allowing sequences module $5$ to which $q^8g(q^{10},q^{25})$ does not contribute, we now avoid contributions from terms of the form $q^{d_r}m(q^{Ma_r},q^{M^2b_r},-1)$.  Here $a_r$, $b_r$, and $d_r$ are integers.  By definition (\ref{equation:mxqz-def}) the Appell--Lech sums contribute only powers of $q^{M}$ to the Fourier expansion, so we just need to avoid $q$-exponents congruent to $d_r$ modulo $M$.

We list a few more facts which will be useful in the paper.  We note the $z=x^{-1}$ and $z=-x^{-3}$ specialisations of identity (\ref{equation:g-full}):
\begin{subequations}
\begin{equation}
g(x,q)=-x^{-1}m(q^2x^{-3},q^3,x^2)-x^{-2}m(qx^{-3},q^3,x^2),\label{equation:g-to-m}
\end{equation}
\begin{equation}
g(x,q)=-x^{-2}m(qx^{-3},q^3,-1)-x^{-1}m(q^2x^{-3},q^3,-1)+\Delta(x;q),\label{equation:g-spec1}
\end{equation}
\end{subequations}
where
\begin{align}
\Delta(x;q):=\frac{x^{-2}J_1J_{3}^3j(-x^{2};q)}{j(x;q)j(-qx^{3};q^3)j(-q^{2}x^{3};q^3)\overline{J}_{0,3}}.\label{equation:theta-delta}
\end{align}
For computing examples involving $g(x,q)$, we recall an identity {\cite[p. $32$]{RLN}, \cite[$(12.5.3)$]{ABI}}:
\begin{equation}
g(x,q)=-x^{-1}+qx^{-3}g(-qx^{-2},q^4)-qg(-qx^2,q^4)+\frac{J_2J_{2,4}^2}{xj(x;q)j(-qx^2;q^2)},\label{equation:g-split}
\end{equation}
as well as two properties:
\begin{align}
g(x,q)=g(q/x,q) {\text{ and }}  g(qx,q)=-x-x^2-x^3g(x,q).
\end{align}

\section{The Main Theorem}\label{section:theorem}
\begin{theorem}\label{theorem:Theorem}
If $0\leq a<M$, then
\begin{equation*}
D(a, M,q) = d(a, M,q) + T_{a, M}(q)
\end{equation*}
where
\newpage

\enlargethispage{3\baselineskip}
\begin{equation*}
d(a, M,q)=
 \begin{cases}
2\,m((-1)^{M+1}q^{M(M-1)/6},q^{M^2/3},z)
\hspace{.5in}\mathrm{if\ }a=0 \mathrm{\ and\ }M\equiv 0\ (\mod3)\\[.07in]
(-1)^a\ q^{-a(a+1)/6}\ m((-1)^{M+1}\ q^{M(M-2a-1)/6},q^{M^2/3},z)\\
\hspace{1in}+ (-1)^{a+1}\ q^{-a(a-1)/6}\ m((-1)^{M+1}\ q^{M(M-2a+1)/6},q^{M^2/3},z)\\
\hspace{1.5in}\mathrm{if\ }a\equiv 0\ (\mod3) \mathrm{,\ }M\equiv 0\ (\mod3)\mathrm{,\ and\ }a\neq0\\[.07in]
(-1)^a\ q^{-a(a-1)/6}\ m((-1)^{M+1}\ q^{M(M-2a+1)/6},q^{M^2/3},z)\\
\hspace{1.5in}\mathrm{if\ }a\equiv 1\ (\mod3)\mathrm{\ and\ }M\equiv 0\ (\mod3)\\[.07in]
\rule{5in}{.5pt}\\
0\hspace{1.44in}\mathrm{if\ }a=0 \mathrm{\ and\ }M=1\\[.07in]
2+2\ (-1)^{M+1}\ q^{M(M-1)/6}\ g((-1)^{M+1}\ q^{M(M-1)/6},q^{M^2})\\
\hspace{1.5in}\mathrm{if\ }a=0 \mathrm{,\ }M\equiv 1\ (\mod3)\mathrm{,\ and\ }M\neq1\\[.07in]
(-1)^a q^{-a(a+1)/6}
 +(-1)^{M+a+1}\ q^{(M(M-2a-1)-a(a+1))/6}\ g((-1)^{M+1} q^{M(M-2a-1)/6},q^{M^2})\\
\hspace{1in}+\;(-1)^{M+a+1}\ q^{(M(M+2a-1)-a(a-1))/6}\ g((-1)^{M+1} q^{M(M+2a-1)/6},q^{M^2})\\
\hspace{1.5in}\mathrm{if\ }a\equiv 0\ (\mod3)\mathrm{,\ }M\equiv 1\ (\mod3) \mathrm{,\ }a\neq0 \mathrm{,\ and\ }M\neq 2a+1\\[.07in]
(-1)^a q^{a(7a+5)/6}\ g(q^{2aM/3},q^{M^2})
\hspace{.5in}\mathrm{if\ }a\equiv 0\ (\mod3)\mathrm{,\ }a\neq0 \mathrm{,\ and\ }M=2a+1\\[.07in]

(-1)^{M+a+1}\ q^{(M(M+2a+1)-a(a+1))/6}\ g((-1)^{M+1}\ q^{M(M+2a+1)/6},q^{M^2})\\
\hspace{1in}+ (-1)^{a+1}\ q^{(2M^2-a(a-1))/6}\ g((-1)^{M+1}\ q^{M(3M+2a-1)/6},q^{M^2})\\
\hspace{1.5in}\mathrm{if\ }a\equiv 2\ (\mod3)\mathrm{\ and\ }M\equiv 1\ (\mod3)\\
\rule{5in}{.5pt}\\
2\ (-1)^M\ q^{M(M+1)/6}\ g((-1)^{M+1}\ q^{M(M+1)/6},q^{M^2})\\
\hspace{1.5in}\mathrm{if\ }a=0 \mathrm{\ and\ }M\equiv 2\ (\mod3)\\[.07in]
(-1)^{a+1}\ q^{-a(a-1)/6}
 + (-1)^{M+a}\ q^{(M(M-2a+1)-a(a-1))/6}\ g((-1)^{M+1}\ q^{M(M-2a+1)/6},q^{M^2})\\
\hspace{1.2in}+\;(-1)^{M+a}\ q^{(M(M+2a+1)-a(a+1))/6}\ g((-1)^{M+1}\ q^{M(M+2a+1)/6},q^{M^2})\\
\hspace{1.5in}\mathrm{if\ }a\equiv 0\ (\mod3) \mathrm{,\ }M\equiv 2\ (\mod3) \mathrm{,\ }a\neq0\mathrm{,\ and\ }M\neq 2a-1\\[.07in]
(-1)^{a+1}\ q^{a(7a-5)/6}\ g(q^{2aM/3},q^{M^2})
\hspace{.5in}\mathrm{if\ }a\equiv 0\ (\mod3)\mathrm{\ and\ }M=2a-1\\
(-1)^{M+a}\ q^{(M(M+2a-1)-a(a-1))/6}\ g((-1)^{M+1}\ q^{M(M+2a-1)/6},q^{M^2})\\
\hspace{1in}+ (-1)^a\ q^{(2M^2-a(a+1))/6}\ g((-1)^{M+1}\ q^{M(3M-2a-1)/6},q^{M^2})\\
\hspace{1.5in}\mathrm{if\ }a\equiv 1\ (\mod3)\mathrm{\ and\ }M\equiv 2\ (\mod3)
\end{cases}
\end{equation*}
and $T_{a,M}(q)$ is a theta function.   For $M\equiv 0 \pmod 3$, $T_{a,M}(q)$ also depends on the $z$.
\end{theorem}

We have removed the cases determined by $D(a,M)=D(M-a,M).$  We note that the case $a=0$ is almost a special case of $a\equiv0 \pmod3$; we just have to add $1$.  This is related to the empty partition of $0$, whose rank Atkin and Swinnerton-Dyer leave undefined.  We have defined it to have rank $0$.

For a theta function $f(q)$ (and for many other functions) there is an associated fractional exponent $b/a$ such that $q^{b/a}\cdot f(q)$ is, in some ways, simpler than $f(q).$  Modular properties of theta functions are easier to state when you multiply by $q^{b/a}.$  Let us call $b/a$ the multiplier exponent or $\lambda$.  The $\lambda$ for both $J_{a,m}$ and $\overline{J}_{a,m}$ is $(m-2a)^2/8m.$  For $m(q^a,q^m,z)$, $\lambda$ is $-(m-2a)^2/8m.$  We note that the $\lambda$'s for all of the cases and summands therein of the theorem are equal to $-1/24.$  When multiplied by $q^{-1/24}$,  the term $T_{a,M}(q)$ becomes a weight $1/2$ weakly holomorphic modular form.

 In Section \ref{section:proof},  we see that we can write $T_{a,M}(q)$ in terms of the theta function $j(x;q)$.   We explain the arbitrary $z$ as well as how and when $T_{a,M}(q)$ depends on it.  When using Theorem \ref{theorem:rootsof1}, we have an arbitrary $z^{\prime}$ in both the resulting $m(x,q,z)$ function and the quotients of theta functions.  For $M\equiv 1,  \ 2 \pmod{3}$ we specialise the $z^{\prime}$ in order to form $g(x,q)$ terms as in (\ref{equation:g-full}).  For $M\equiv 0 \pmod 3$, we cannot obtain $g(x,q)$ terms, so we leave the $z^{\prime}$ arbitrary for the reader to decide; this situation is not unlike identity (\ref{equation:g-full}) in which the right-hand side is $z$-independent.

\section{Examples of Formulas for $D(a,M)$}
We state formulas for small values of $M$ and give examples of proofs for some of the individual $D(a,M,q)$'s.
\subsection{ Case {M=2}:}
\begin{align}
D(0,2)&=2qg(-q,q^4) + \frac{1}{2}\cdot \frac{J_{1,2}^2}{ J_{1}}\\
D(1,2)&=-D(0,2)=-2qg(-q,q^4) - \frac{1}{2}\cdot \frac{J_{1,2}^2}{  J_{1}}
\end{align}

To prove $D(0,2)$, we use (\ref{equation:g-split}) to obtain
\begin{align*}
g(-1,q)&=1-2qg(-q,q^4)-\frac{1}{2}\cdot \frac{J_1^3}{J_2^2}.
\end{align*}
Using (\ref{equation:D-to-g}) we arrive at
\begin{align*}
D(0,2)&=\frac{1}{2}\cdot 1\cdot(1-(-1))\cdot (1-g(-1,q))=2qg(-q,q^4)+\frac{1}{2}\cdot \frac{J_1^3}{J_2^2}.
\end{align*}

\subsection{Case M=3:}
\begin{align}
D(0,3)&=2m(q,q^3,-1) - \frac{1}{3}\cdot \frac{J_{1,2}^2 }{\overline{J}_{1,3}}\\
D(1,3)&=D(2,3) = -D(0,3)/2 = -m(q,q^3,-1) + \frac{1}{6}\cdot \frac{J_{1,2}^2}{\overline{J}_{1,3}}
\end{align}

For $D(0,3)$, we specialize $M=3$ and $z^{\prime}=-1$ in (\ref{equation:Dfull-a-equal-0-and-3-div-M}) and note straightforward facts such as $j(\zeta_3;q)=(1-\zeta_3)J_3$ and $j(-\zeta_3;q)=(1+\zeta_3)J_1^2J_6/J_2J_3$. It follows that
\begin{align*}
D(0,3)&=2m(q,q^3,-1)+\frac{1}{3}\frac{J_1^2}{\overline{J}_{0,1}}\Big [ \frac{(\zeta_3-\zeta_3^2)}{(1-\zeta_3)J_3}\frac{(1+\zeta_3)J_1^2J_6}{J_2J_3}
+  \frac{(\zeta_3^2-\zeta_3)}{(1-\zeta_3^2)J_3}\frac{(1+\zeta_3^2)J_1^2J_6}{J_2J_3}  \Big ]\\
&=2m(q,q^3,-1)-\frac{J_1^5J_6}{3J_2^3J_3^2},
\end{align*}
where the last line follows from  the fact $1+\zeta_3+\zeta_3^2=0$ and some simplifying.

\subsection{Case M=7:}
{\allowdisplaybreaks \begin{align}
&D(0,7) = 2 + 2 q^7 g(q^7, q^{49})\\
&         - \frac{8}{7}\cdot \frac{     J_{21,49}^2 }{ J_7}
+ \frac{6}{7}\cdot  q^1 \cdot \frac{ J_{49}^2}{  J_{7,49}}
- \frac{2}{7}\cdot q^2 \cdot \frac{J_{14,49}^2}{ J_{7}}
+ \frac{4}{7}\cdot  q^3 \cdot \frac{J_{49}^2 }{ J_{14,49}}
+ \frac{2}{7}\cdot q^4 \cdot \frac{J_{49}^2 }{ J_{21,49}}
- \frac{4}{7}\cdot q^6\cdot \frac{ J_{7,49}^2 }{ J_{7}}\notag\\
&D(1,7)= D(6,7) = -1 - q^7 g(q^7, q^{49}) + q^{16} g(q^{21}, q^{49})\label{equation:D17}\\
&  + \frac{6}{7}\cdot \frac{J_{21,49}^2}{ J_{7}}
- \frac{1}{7}\cdot q^1 \cdot \frac{J_{49}^2}{  J_{7,49}}
+ \frac{5}{7}\cdot q^2 \cdot \frac{J_{14,49}^2}{ J_{7}}
- \frac{3}{7}\cdot q^3 \cdot \frac{J_{49}^2 }{ J_{14,49}}
+ \frac{2}{7}\cdot q^4 \cdot \frac{J_{49}^2}{ J_{21,49}}
+ \frac{3}{7}\cdot q^6 \cdot \frac{J_{7,49}^2}{ J_7}\notag\\
&D(2,7) = D(5,7) = q^{13} g(q^{14}, q^{49}) - q^{16} g(q^{21}, q^{49})\\
& - \frac{1}{7} \cdot \frac{J_{21,49}^2}{ J_7}
- \frac{1}{7}\cdot  q^1 \cdot \frac{  J_{49}^2}{  J_{7,49}}
- \frac{2}{7}\cdot  q^2 \cdot \frac{J_{14,49}^2}{ J_{7}}
+ \frac{4}{7} \cdot q^3 \cdot \frac{J_{49}^2}{  J_{14,49}}
- \frac{5}{7}\cdot q^4 \cdot \frac{J_{49}^2}{  J_{21,49}}
+ \frac{3}{7} \cdot q^6 \cdot \frac{J_{7,49}^2}{ J_7}\notag\\
&D(3,7) = D(4,7) = - q^{13} g(q^{14}, q^{49})\\
&   - \frac{1}{7}\cdot \frac{     J_{21, 49}^2}{ J_{7}}
- \frac{1}{7}\cdot  q^1 \cdot \frac{   J_{49}^2 }{ J_{7, 49}}
- \frac{2}{7}\cdot  q^2 \cdot \frac {J_{14, 49}^2}{  J_{7}}
- \frac{3}{7}\cdot  q^3 \cdot \frac{J_{49}^2}{ J_{14, 49}}
+ \frac{2}{7}\cdot  q^4 \cdot \frac{J_{49}^2}{ J_{21, 49}}
- \frac{4}{7}\cdot  q^6 \cdot \frac{J_{7, 49}^2}{ J_{7}}\notag
\end{align}}%

Dyson's (\ref{equation:ASD-2.5})--(\ref{equation:ASD-2.11}) and Atkin and Swinnerton-Dyer's \cite[Theorem $5$]{ASD} are immediate. Dyson's conjecture (\ref{equation:Dyson-M7}) is clear as there are no terms of the form $q^{7n+5}$.  Considering coefficients of respective $g(x,q)$ terms easily leads to identities such as (\ref{equation:ASD-2.6}).  In particular, the $g$ expressions in $D(1,7)$ only contribute to the $q^{7n}$ and $q^{7n+2}$ progressions, so for the $q^{7n+1}$ terms we only have
\begin{equation*}
- \frac{1}{7}\cdot  q^1 \cdot \frac{  J_{49}^2}{  J_{7,49}}.
\end{equation*}

\section{Proof of the Main Theorem}\label{section:proof}

The proofs of the twelve cases are similar, so we give only a few as examples.  For the sake of brevity, we do not write out an explicit theta function for each $T_{a,M}(q)$.  However, we point out that our arguments are effective in that we can easily keep track of the various summands of quotients of theta functions that arise from using (\ref{equation:m-change-z}) and Theorem \ref{theorem:rootsof1}.

We recall that $\zeta_M:=e^{2\pi i /M}$, so by (\ref{equation:D-to-g}):
{\allowdisplaybreaks \begin{align}
&\sum_{n\geq 0}\left(N(a,M,n) - \frac{p(n)}{M}\right) q^n =\frac{1}{M}\sum_{j=1}^{M-1}\zeta_M^{-aj}\Big (1-\zeta_M^j\Big )\Big (1+\zeta_M^jg(\zeta_M^j,q)\Big )\notag\\
&=\frac{1}{M}\sum_{j=1}^{M-1}\zeta_M^{-aj}\Big (1-\zeta_M^j\Big )\Big (1-m\big (q^2\zeta_M^{-3j},q^3,-1\big )-\zeta_M^{-j}m\big (q\zeta_M^{-3j},q^3,-1\big ) + \zeta_M^j\Delta(\zeta_M^j;q) \Big )\notag\\
&= \frac{1}{M}\sum_{j=0}^{M-1}\zeta_M^{-aj}\Big (m\big (q\zeta_M^{3j},q^3,-1\big )-\zeta_M^{-j}m\big (q\zeta_M^{-3j},q^3,-1\big )
 -\zeta_M^j m\big (q\zeta_M^{3j},q^3,-1\big )\notag \\
 &\ \ \ \ \ \ \ \ \ \ +m\big (q\zeta_M^{-3j},q^3,-1\big )\Big )  +\frac{1}{M}\sum_{j=1}^{M-1}\zeta_M^{-aj}\Big (1-\zeta_M^j\Big ) \zeta_M^j\Delta(\zeta_M^j;q),\label{equation:DAMQ-heuristic}
\end{align}}%
where the second equality follows from (\ref{equation:g-spec1}) and the third equality follows from (\ref{equation:m-fnq-flip}) and (\ref{equation:m-fnq-x}).  Note the change in the lower limit of the first summation symbol in the last equality.

We consider the case $a=0$, $M\equiv 0 \pmod 3$, and recall that $T_{a,M}(q)$ will depend on an arbitrary $z$.  We write $j=t(M/3)+r$, where $0\le t\le 2$ and $0\le r\le (M/3)-1.$  The first summand in $(\ref{equation:DAMQ-heuristic})$ is then
{\allowdisplaybreaks \begin{align}
\frac{1}{M}&\sum_{j=0}^{M-1}m\big (q\zeta_M^{3j},q^3,-1\big )=\frac{1}{M}\sum_{t=0}^2\sum_{r=0}^{(M/3)-1}m\big (q\zeta_{M/3}^{t(M/3)+r},q^3,-1\big )\notag \\
&=\frac{3}{M}\sum_{r=0}^{(M/3)-1}m\big (q\zeta_{M/3}^{r},q^3,-1\big )\notag \\
&= m\big ( (-1)^{M+1} q^{M(M-1)/6}, q^{M^2/3}, z^{\prime} \big ) +\Psi_0^{M/3}(q,-1,z^{\prime};q^3),\label{equation:theta-last}
\end{align}}%
where the last equality follows from Theorem \ref{theorem:rootsof1} with $k=0$, $n=M/3$, $z=-1$.  The fourth summand is similar.  For the second summand,
{\allowdisplaybreaks \begin{align}
-\frac{1}{M}\sum_{j=0}^{M-1}\zeta_M^{-j}m\big (q\zeta_M^{-3j},q^3,-1\big )&=-\frac{1}{M}\sum_{t=0}^2\sum_{r=0}^{(M/3)-1}\zeta_M^{-(t(M/3)+r)}m\big (q\zeta_{M/3}^{-t(M/3)-r},q^3,-1\big )\notag \\
&=-\frac{1}{M}\Big ( \sum_{t=0}^2 \zeta_3^{-t}\Big )\sum_{r=0}^{(M/3)-1}\zeta_M^{-r}m\big (q\zeta_{M/3}^{-r},q^3,-1\big )= 0.
\end{align}}%
The third summand is similar and the result follows. So for $M\equiv 0 \pmod 3$ and $a=0$:
{\allowdisplaybreaks \begin{align}
D&(0,M,q)= 2m\big ( (-1)^{M+1} q^{M(M-1)/6}, q^{M^2/3}, z^{\prime} \big ) \notag\\
&\ \ \ \ \ +2\cdot \frac{ J_{M^2/3}^3} {\overline{J}_{0,3}}\cdot\sum_{t=0}^{(M/3)-1} \frac{q^{3\binom{t+1}{2}} j\big(-q^{3\binom{M/3+1}{2}+tM}/z^{\prime};q^{M^2/3}\big)
j\big ( (-1)^{M/3}q^{Mt+M/3} z^{\prime};q^{M^2/3}\big )}
{j\big( z^{\prime},(-1)^{M/3+1}q^{3\binom{M/3}{2}+M/3} z^{\prime}, (-1)^{M/3}q^{Mt+M/3};q^{M^2/3}\big)}\notag \\
&\ \ \ \ \ +\frac{1}{M}\sum_{j=1}^{M-1}\Big (1-\zeta_M^j\Big )  \frac{\zeta_M^{-j}J_1J_{3}^3j(-\zeta_M^{2j};q)}{j(\zeta_M^j;q)j(-q\zeta_M^{3j};q^3)j(-q^{2}\zeta_M^{3j};q^3)\overline{J}_{0,3}},\label{equation:Dfull-a-equal-0-and-3-div-M}
\end{align}}%
where the theta quotients are from (\ref{equation:theta-last}) using (\ref{equation:msplit-theta}), and (\ref{equation:DAMQ-heuristic}) using (\ref{equation:theta-delta}).

We consider the case $a\equiv 1 \pmod 3$, $M\equiv 1 \pmod 3$, $M\neq 2a-1.$  Here $a=3b+1$, $M=3R+1$, so $a\equiv 3b-3R \pmod M$.  The first summand in $(\ref{equation:DAMQ-heuristic})$ is then
\begin{align}
\frac{1}{M}&\sum_{j=0}^{M-1}\zeta_M^{-aj} m\big (q\zeta_M^{3j},q^3,-1\big )
=\frac{1}{M}\sum_{j=0}^{M-1}\zeta_M^{3(\frac{M-1}{3}-\frac{a-1}{3})j}m\big (q\zeta_{M}^{3j},q^3,-1\big )\notag\\
&= (-1)^aq^{-a(a+1)/6-2M^2/3-2Ma/3-M/3}m\big ( (-1)^{M+1} q^{-M^2/2-Ma-M/2}, q^{3M^2}, -1 \big )\notag\\
&\ \ \ \ \  +\Psi_{(2M+a)/3}^M(q,-1,-1;q^3)\notag \\
&= (-1)^{M+a+1}q^{-a(a+1)/6-M^2/6+Ma/3+M/6} m\big ( (-1)^{M+1} q^{2M^2-M(3M-2a-1)/2}, q^{3M^2}, -1 \big )\notag \\
&\ \ \ \ \ +\Psi_{(2M+a)/3}^M(q,-1,-1;q^3), \label{equation:sumA}
\end{align}
where the second equality follows from Theorem \ref{theorem:rootsof1} with $k=(2M+a)/3$ and the third equality follows from (\ref{equation:m-fnq-flip}).  For the second summand in (\ref{equation:DAMQ-heuristic}) we write $a=3b+1$, $M=3R-2$, so $a+1\equiv 3b+2\equiv 3b+3R \pmod M$.  Here
\begin{align}
-\frac{1}{M}&\sum_{j=0}^{M-1}\zeta_M^{-(a+1)j} m\big (q\zeta_M^{-3j},q^3,-1\big )
=-\frac{1}{M}\sum_{j=0}^{M-1}\zeta_M^{-3(\frac{a-1}{3}+\frac{M+2}{3})j}m\big (q\zeta_{M}^{-3j},q^3,-1\big )\notag\\
&= (-1)^aq^{-a(a+1)/6-2M^2/3+2Ma/3+M/3}m\big ( (-1)^{M+1} q^{M^2-M(3M-2a-1)/2}, q^{3M^2}, -1 \big )\label{equation:sumB} \\
&\ \ \ \ \ -\Psi_{(2M-a-1)/3}^M(q,-1,-1;q^3), \notag
\end{align}
where the second equality follows from Theorem \ref{theorem:rootsof1} with $k=(2M-a-1)/3$.
For the third summand in (\ref{equation:DAMQ-heuristic}) we write $a=3b+1$.  Here
{\allowdisplaybreaks \begin{align}
-\frac{1}{M}&\sum_{j=0}^{M-1}\zeta_M^{-(a-1)j} m\big (q\zeta_M^{3j},q^3,-1\big )
=-\frac{1}{M}\sum_{j=0}^{M-1}\zeta_M^{-3(\frac{a-1}{3})j}m\big (q\zeta_{M}^{3j},q^3,-1\big )\notag\\
&= (-1)^aq^{-a(a-1)/6}m\big ( (-1)^{M+1} q^{M^2+M(M-2a+1)/2}, q^{3M^2}, -1 \big )\notag \\
&\ \ \ \ \ -\Psi_{(a-1)/3}^M(q,-1,-1;q^3) \notag \\
&= (-1)^aq^{-a(a-1)/6}-(-1)^aq^{-a(a-1)/6}m\big ( (-1)^{M+1} q^{2M^2-M(M-2a+1)/2}, q^{3M^2}, -1\big )\label{equation:sumC}\\
&\ \ \ \ \ -\Psi_{(a-1)/3}^M(q,-1,-1;q^3), \notag
\end{align}}%
where the second equality follows from Theorem \ref{theorem:rootsof1} with $k=(a-1)/3$ and the third equality follows from (\ref{equation:m-fnq-x}) and then (\ref{equation:m-fnq-flip}).   For the fourth summand in (\ref{equation:DAMQ-heuristic}), we use $k=(M-a)/3$ to obtain
{\allowdisplaybreaks \begin{align}
\frac{1}{M}&\sum_{j=0}^{M-1}\zeta_M^{-aj} m\big (q\zeta_M^{-3j},q^3,-1\big )\notag \\
&= (-1)^{M+a}q^{-a(a-1)/6-M(M-2a+1)/6}m\big ( (-1)^{M+1} q^{M^2-M(M-2a+1)/2}, q^{3M^2}, -1 \big )\label{equation:sumD}\\
&\ \ \ \ \  +\Psi_{(M-a)/3}^M(q,-1,-1;q^3) \notag
\end{align}}%
With (\ref{equation:g-spec1}) in mind, combining (\ref{equation:sumA}) and (\ref{equation:sumB}) yields
{\allowdisplaybreaks \begin{align}
&(-1)^{a+1}q^{(2M^2-a(a+1))/6}g\big ( (-1)^{M+1} q^{M(3M-2a-1)/6},q^{M^2}\big ) \label{equation:finalA}\\
&\ \ \ \ \  -(-1)^{a+1}q^{(2M^2-a(a+1))/6}\Delta((-1)^{M+1}q^{M(3M-2a-1)/6};q^{M^2})\notag\\
&\ \ \ \ \ +\Psi_{(2M+a)/3}^M(q,-1,-1;q^3)-\Psi_{(2M-a-1)/3}^M(q,-1,-1;q^3). \notag
\end{align}}%
Combining (\ref{equation:sumC}) and (\ref{equation:sumD}) and using (\ref{equation:g-spec1}) yields
{\allowdisplaybreaks \begin{align}
 (-1)^{a}&q^{-a(a-1)/6}+(-1)^{M+a+1}q^{(M(M-2a+1)-a(a-1))/6}g\big ( (-1)^{M+1} q^{M(M-2a+1)/6},q^{M^2}\big )\label{equation:finalB}\\
 & -(-1)^{M+a+1}q^{(M(M-2a+1)-a(a-1))/6}\Delta((-1)^{M+1}q^{M(M-2a+1)/6};q^{M^2})\notag \\
 & -\Psi_{(a-1)/3}^M(q,-1,-1;q^3)  +\Psi_{(M-a)/3}^M(q,-1,-1;q^3). \notag
\end{align}}%
Combining (\ref{equation:finalA}) and (\ref{equation:finalB}) gives
\begin{align}
D(a,M,q)&=(-1)^{a}q^{-a(a-1)/6}\notag\\
&\ \ \ +(-1)^{M+a+1}q^{(M(M-2a+1)-a(a-1))/6}g\big ( (-1)^{M+1} q^{M(M-2a+1)/6},q^{M^2}\big )\notag \\
&\ \ \ +(-1)^{a+1}q^{(2M^2-a(a+1))/6}g\big ( (-1)^{M+1} q^{M(3M-2a-1)/6},q^{M^2}\big )+T_{a,M}(q),\notag
\end{align}
where
{\allowdisplaybreaks \begin{align}
T_{a,M}(q)&= -(-1)^{a+1}q^{(2M^2-a(a+1))/6}\Delta((-1)^{M+1}q^{M(3M-2a-1)/6};q^{M^2})\\
 & \ \ \ \ \ -(-1)^{M+a+1}q^{(M(M-2a+1)-a(a-1))/6}\Delta((-1)^{M+1}q^{M(M-2a+1)/6};q^{M^2})\notag \\
 &\ \ \ \ \ +\Psi_{(2M+a)/3}^M(q,-1,-1;q^3)-\Psi_{(2M-a-1)/3}^M(q,-1,-1;q^3) \notag\\
 & \ \ \ \ \ -\Psi_{(a-1)/3}^M(q,-1,-1;q^3)  +\Psi_{(M-a)/3}^M(q,-1,-1;q^3) \notag\\
 &\ \ \ \ \ +\frac{1}{M}\sum_{j=1}^{M-1}\zeta_M^{-aj}\Big (1-\zeta_M^j\Big ) \zeta_M^j\Delta(\zeta_M^j;q).\notag
\end{align} }%
Setting up for (\ref{equation:g-to-m}) instead of (\ref{equation:g-spec1}) in (\ref{equation:sumA})--(\ref{equation:sumD}) gives an alternate form:
\begin{align}
T_{a,M}(q)&=\Psi_{(2M+a)/3}^M(q,-1,q^{-M(3M-2a-1)/3};q^3)-\Psi_{(2M-a-1)/3}^M(q,-1,q^{M(3M-2a-1)/3};q^3) \notag\\
 & \ \ \ \ \ -\Psi_{(a-1)/3}^M(q,-1,q^{-M(M-2a+1)/3};q^3)  +\Psi_{(M-a)/3}^M(q,-1,q^{M(M-2a+1)/3};q^3) \notag\\
 &\ \ \ \ \ +\frac{1}{M}\sum_{j=1}^{M-1}\zeta_M^{-aj}\Big (1-\zeta_M^j\Big ) \zeta_M^j\Delta(\zeta_M^j;q).
\end{align}

\section{On Generalisations of Dyson's Rank-Differences}\label{section:BrO}

In \cite[Sections $7$, $5$]{Za}, Zagier obtains a statement slightly more general than \cite[Theorem $1.4$]{BrO2}.  He states that for all $M>0$ and all $a\in \mathbb{Z}/M\mathbb{Z}$ that the function (\ref{equation:Zag-def}) is a weight $1/2$ weakly holomorphic modular form for any arithmetic progression $\mathcal{A}\subset \mathbb{Z}$ not containing any number of the form $(1-h^2)/24$ with $h\equiv 2a\pm 1 \pmod{2M}$.  Zagier points out that this holds if $\mathcal{A}$ is the set of $n$  with $\Big ( \frac{1-24n}{\mathcal{P}}\Big )=-1$ for some prime $\mathcal{P}>3$, and in this case $\mathcal{A}$ is actually supported on a collection of arithmetic progressions.

It is straightforward to read such results from Theorem \ref{theorem:Theorem}.   We first focus on \cite[Theorem $1.4$]{BrO2}.  Take the case $M\equiv 0 \pmod 3$, $a\equiv 2 \pmod 3$ of Theorem \ref{theorem:Theorem}:
\begin{equation}
D(a,M,q)= (-1)^{a+1}\ q^{-a(a+1)/6}\ m((-1)^{M+1}\ q^{M(M-2a-1)/6},q^{M^2/3},z)+T_{a,M}(q),
\end{equation}
where $T_{a,M}(q)$ is a theta function which depends on the arbitrary $z$.  Using Theorem \ref{theorem:msplit-general-n} with $n=\mathcal{P}$ and $z'=-1$ (note that $n$ is from Theorem \ref{theorem:msplit-general-n} and not an $n\in\mathcal{A}$), we have
\begin{align}
D(a,M,q)&= (-1)^{a+1}\ q^{-a(a+1)/6} \sum_{r=0}^{\mathcal{P}-1} q^{-\frac{M^2}{3}\binom{r+1}{2}}\Big ( (-1)^{M}q^{\frac{M(M-2a-1)}{6}}\Big )^r \cdot  \notag \\
&\ \ \ \ \  \cdot \Big [ m(- q^{\frac{M^2}{3}\big (\binom{\mathcal{P}}{2}-\mathcal{P}r\big )}(  (-1)^{M}\ q^{\frac{M(M-2a-1)}{6}}  )^{\mathcal{P}},q^{\frac{(M\mathcal{P})^2}{3}},-1)\Big ]+T_{a,M}^{\prime}(q),
\end{align}
where $T_{a,M}^{\prime}(q)$ is $T_{a,M}(q)$ after having absorbed the theta functions which appear when using Theorem \ref{theorem:msplit-general-n}.  With the Appell--Lerch sum definition (\ref{equation:mxqz-def}) in mind, we can write
\begin{align}
D(a,M,q)&= (-1)^{Mr+a+1}\ q^{-a(a+1)/6} \sum_{r=0}^{\mathcal{P}-1} q^{-\frac{M^2}{3}\binom{r+1}{2}+\frac{rM(M-2a-1)}{6}}\sum_{m=m_r}^{\infty}a_r(m)q^{\mathcal{P}m} +T_{a,M}^{\prime}(q).\label{equation:mockness}
\end{align}
Multiplying by  $q^{-1/24}$ makes $T_{a,M}^{\prime}(q)$ a weight $1/2$ weakly holomorphic modular form.   We again remind the reader of the  various ways of producing new modular forms from old ones such as taking the twist by a Dirichlet character or restricting the index of summation of the Fourier expansion to terms that lie in certain arithmetic progressions.    So to ensure modularity, we only want the terms determined by the set $\mathcal{A}$ to be contributed from $T_{a,M}^{\prime}(q)$; in other words, we only have to worry about picking up `mockness'  from the $m(x,q,z)$ portion in (\ref{equation:mockness}) when we have for some $n\in \mathcal{A}$:
\begin{equation}
-\frac{a(a+1)}{6}-\frac{M^2}{3}\binom{r+1}{2}+\frac{rM(M-2a-1)}{6}\equiv n \pmod {\mathcal {P}}.\label{equation:cond-1}
\end{equation}
Condition (\ref{equation:cond-1}) implies
\begin{equation}
1-24n\equiv (2rM+2a+1)^2 \pmod{\mathcal{P}}.
\end{equation}
The other subcases for the case $M\equiv 0 \pmod 3$ imply
\begin{equation}
1-24n\equiv (2rM+2a\pm 1)^2 \pmod{\mathcal{P}},
\end{equation}
where the sign depends on $a$.  So when
\begin{equation}
\Big ( \frac{1-24n}{\mathcal{P}}\Big )=-1,
 \end{equation}
 then we have nothing to worry about.  The other cases $M\equiv 1,2\pmod 3$ are similar.  Here one needs Theorem  \ref{theorem:msplit-general-n} as well as (\ref{equation:g-full}).

We consider \cite[Theorem $1.1$]{BrOR}.  Here, $M\ge 5$ is prime.  For a given summand $q^mg(q^{Mk},q^{M^2})$, if $m\equiv d \pmod M$, it is easy to check that $1-24d\equiv (2a\pm 1)^2 \pmod {2M}$ where the sign depends on $M$ and $a.$  For example, in the first summand in the case $M\equiv 2 \pmod 3$, $a\equiv 1 \pmod 3$, the condition
\begin{equation}
\frac{M(M+2a-1)-a(a-1)}{6}\equiv d \pmod M
\end{equation}
implies
\begin{equation}
1-24d\equiv (2a-1)^2 \pmod {2M}.\label{equation:careful}
\end{equation}
If $\big ( \frac{1-24d}{M} \big ) =-1$,  we will never have an $m\equiv d \pmod M$, so any rank-difference can be taken.  If $\big ( \frac{1-24d}{M} \big ) =1$, then we may encounter $m$'s with $m\equiv d \pmod M$,  so one must be careful to avoid the rank values $a$ determined by (\ref{equation:careful}).

\section{On $M_2$ rank-differences for partitions without repeated odd parts}\label{section:M2}

There are also variations of rank-differences.  Take for example the $M_2$ rank of Berkovitch and Garvan \cite{BG}, which is based on work of Dyson \cite{D2,D3}.   Here
\begin{align*}
M_2-\textup{rank}(\lambda)&:=\Big \lceil \tfrac{\ell (\lambda)}{2}\Big \rceil-\nu(\lambda)
\end{align*}
where $\ell (\lambda)$ is the largest part of $\lambda$ and $\nu(\lambda)$ is the number of parts of $\lambda$.  The $M_2$ rank of a partition is also the number of columns minus the number of rows in the $2$-modular diagram \cite{LO}, where, for example the partition $\lambda=(10,7,2,1)$ has $2$-modular diagram:
\[ \begin{array}{ccccc}
2&2&2&2&2\\
2&2&2&1\\
2\\
1
\end{array}
\]
Similar to the rank-differences of Dyson \cite{D1} and Atkin and Swinnerton-Dyer \cite{ASD}, Lovejoy and Osburn \cite[Theorems $1.1$, $1.2$]{LO} proved results on $M_2$ rank-differences for partitions without repeated odd parts.

 Let $N_2(m,n)$ denote the number of partitions of $n$ without repeated odd parts whose $M_2$ rank is $m$.  We have the following nice generating function \cite[$(1.1)$]{LO}:
\begin{equation}
\sum_{\substack{n\ge0 \\ m\in \mathbb{Z}}}N_2(m,n)z^mq^n=\sum_{n=0}^{\infty}\frac{q^{n^2}(-q;q^2)_n}{(zq^2,q^2/z;q^2)_n}.
\end{equation}
Define $N_2(a,M;n)$ to be the number of partitions of $n$ without repeated odd parts whose $M_2$ rank is congruent to $a$ mod $M$, and define $p_o(n)$ to be the number of partitions of $n$ without repeated odd parts.   Furthermore we define the $M_2$ rank-difference
\begin{equation}
R_2(a,b,M,c,m):=\sum_{n= 0}^{\infty} \Big ( N_2(a,M;mn+c)-N_2(b,M;mn+c)\Big )q^n.\label{equation:R2-def}
\end{equation}

Lovejoy and Osburn showed \cite[Theorem $1.1$]{LO} for $M=m=3$, slightly rewritten:
{\begin{subequations}
\begin{equation}
R_2(0,1,3,0,3)=-1+3m(q^{15},q^{36},q^9)+3q^{-6}m(q^{-3},q^{36},q^{27})+{\overline{J}_{3,6}^2J_{1,4}}/{J_2^2}\label{equation:R_2-A}
\end{equation}
\begin{equation}
R_2(0,1,3,1,3)={\overline{J}_{3,6}J_{3,12}}/{J_2}\label{equation:R_2-B}
\end{equation}
\begin{equation}
R_2(0,1,3,2,3)={J_3J_{12}^2}/{J_4J_{1,6}}\label{equation:R_2-C}
\end{equation}
\end{subequations}}
We briefly sketch how we prove results for $M_2$ rank-differences similar to our Main Theorem.   We define the following:
\begin{equation}
D_2(a,M):=D_2(a,M,q)=\sum_{n=0}^{\infty}\Big ( N_2(a,M;n)-\frac{p_o(n)}{M}\Big )q^n.
\end{equation}
  From \cite{RLN, AM}, \cite[$(2.13)$]{Mo1}:
\begin{align}
\sum_{n= 0}^{\infty}\frac{(-1)^nq^{n^2}(q;q^2)_{n}}{(-x;q^2)_{n+1}(-q^2/x;q^2)_{n}}&=m(x,q,-1)+\frac{J_{1,2}^2}{2j(-x;q)},\label{equation:RLNid2}
\end{align}
so we can write
\begin{equation}
\sum_{\substack{n\ge0 \\ m\in \mathbb{Z}}}N_2(m,n)z^mq^n=\big (1-z\Big )\Big (m(-z,-q,-1) +\frac{\overline{J}_{1,2}}{2j(z,q)}\Big).
\end{equation}
It follows that
\begin{equation}
D_2(a,M,q)=\sum_{j=1}^{M-1}\zeta_M^{-aj}\Big (1-\zeta_M^j\Big )\Big (m(-\zeta_M^j,-q,-1) +\frac{\overline{J}_{1,2}}{2j(\zeta_M^j,q)}\Big).\label{equation:D2-def}
\end{equation}

 Using (\ref{equation:D2-def}) and Theorem \ref{theorem:rootsof1} we obtain
\begin{theorem}\label{theorem:MT} If $0\le a<M$, then
\begin{align*}
D_2(a,M,q)= (-q)^{-\binom{a+1}{2}}m&\Big ( -(-q)^{\binom{M}{2}-Ma},(-q)^{M^2},z_0\Big)\\
&-(-q)^{-\binom{a}{2}}m\Big ( -(-q)^{\binom{M}{2}-M(a-1)},(-q)^{M^2},z_1\Big)+T_{a,M}(q),
\end{align*}
where $T_{a,M}(q)$ is a theta function which depends on $z_0$ and $z_1$.  Here $z_0$ and $z_1$ are arbitrary.
\end{theorem}
When multiplied by $q^{-1/8}$, the term $T_{a,M}(q)$ becomes a weight $1/2$ weakly holomorphic modular form.

Theorem \ref{theorem:MT} is useful for understanding results of Lovejoy and Osburn \cite[Theorems $1.1$, $1.2$]{LO}.   For the case $M=3$ we can use Theorem \ref{theorem:msplit-general-n} with $n=2$ to show
\begin{align*}
D_2(0,3)&=-1+2m(q^{15},q^{36},q^9)+2q^{-6}m(q^{-3},q^{36},q^{27})\\
&\ \ \ \ \ +\frac{2}{3}\cdot \frac{\overline{J}_{9,18}^2J_{3,12}}{J_6^2}+q\cdot \frac{2}{3}\cdot \frac{\overline{J}_{9,18}J_{9,36}}{J_{6}}+q^2\cdot \frac{2}{3}\cdot \frac{J_{9,36}^2}{J_{3,12}},\\
D_2(1,3)&=-m(q^{15},q^{36},q^9)-q^{-6}m(q^{-3},q^{36},q^{27})\\
&\ \ \ \ \ -\frac{1}{3}\cdot \frac{\overline{J}_{9,18}^2J_{3,12}}{J_6^2}-q\cdot \frac{1}{3}\cdot \frac{\overline{J}_{9,18}J_{9,36}}{J_{6}}-q^2\cdot \frac{1}{3}\cdot \frac{J_{9,36}^2}{J_{3,12}}.
\end{align*}
Identities (\ref{equation:R_2-A})--(\ref{equation:R_2-C}) are immediate.  For  $M=5$, we use Theorem \ref{theorem:msplit-general-n} with $n=2$ to show
 \begin{align*}
D_2&(0,5)=-1-2q^{-55}m(q^{-55},q^{100},q^{75})-2q^{-15}m(q^{-5},q^{100},q^{25})\\
&-\frac{6}{5}\cdot \frac{J_{5,50}^2J_{30,100}J_{40,100}}{J_{5}\overline{J}_{50,200}J_{100}}+q\cdot \frac{4}{5}\cdot \frac{J_{50}^4}{J_{10,50}J_{25}J_{100}}
 +q^2\cdot \frac{2}{5}\cdot \Big ( 2\cdot \frac{J_{25}^2J_{30,100}}{J_{10,25}J_{50}}-q^5\frac{J_{10,100}J_{25}^2}{J_{5,25}J_{50}}\Big )\\
 &-q^3\cdot \frac{2}{5}\cdot \frac{\overline{J}_{25,50}J_{25,100}}{J_{20,50}}
 +q^4\cdot \frac{2}{5}\cdot \frac{J_{10}J_{100}J_{15,50}^2}{J_{5}J_{30,100}J_{40,100}},\\
 D_2&(1,5)=-q^{-1}m(q^{35},q^{100},q^{25})-q^{-6}m(q^{15},q^{100},q^{25})\\
 &\ \ \ \ \ \ \ \ \ \  -1+q^{-55}m(q^{-55},q^{100},q^{75})+q^{-15}m(q^{-5},q^{100},q^{25})\\
&+\frac{4}{5}\cdot \frac{J_{5,50}^2J_{30,100}J_{40,100}}{J_{5}\overline{J}_{50,200}J_{100}}-q\cdot \frac{1}{5}\cdot \frac{J_{50}^4}{J_{10,50}J_{25}J_{100}}
 -q^2\cdot \frac{1}{5}\cdot \Big ( \frac{J_{25}^2J_{30,100}}{J_{10,25}J_{50}}-3\cdot q^5\frac{J_{10,100}J_{25}^2}{J_{5,25}J_{50}}\Big )\\
 &+q^3\cdot \frac{3}{5}\cdot \frac{\overline{J}_{25,50}J_{25,100}}{J_{20,50}}
 +q^4\cdot \frac{2}{5}\cdot \frac{J_{10}J_{100}J_{15,50}^2}{J_{5}J_{30,100}J_{40,100}},\\
D_2&(2,5)=q^{-1}m(q^{35},q^{100},q^{25})+q^{-6}m(q^{15},q^{100},q^{25})\\
&-\frac{1}{5}\cdot \frac{J_{5,50}^2J_{30,100}J_{40,100}}{J_{5}\overline{J}_{50,200}J_{100}}-q\cdot \frac{1}{5}\cdot \frac{J_{50}^4}{J_{10,50}J_{25}J_{100}}
 -q^2\cdot \frac{1}{5}\cdot \Big ( \frac{J_{25}^2J_{30,100}}{J_{10,25}J_{50}}+2\cdot q^5\frac{J_{10,100}J_{25}^2}{J_{5,25}J_{50}}\Big )\\
 &-q^3\cdot \frac{2}{5}\cdot \frac{\overline{J}_{25,50}J_{25,100}}{J_{20,50}}
 -q^4\cdot \frac{3}{5}\cdot \frac{J_{10}J_{100}J_{15,50}^2}{J_{5}J_{30,100}J_{40,100}}.
\end{align*}

It is straightforward to make generalisations on $M_2$ ranks similar to \cite[Theorem $1.4$]{BrO2} and \cite[Theorem $1.1$]{BrOR}

redpajama_set_name: RedPajamaArXiv


text:
The Pinacoteca Civica of Forlì, one of the civic museums of Forlì and currently based in the Musei di San Domenico, is an Italian art gallery. Artists whose work the gallery exhibits include:

 Livio Agresti
 Clemente Alberi
 Beato Angelico
 Nicola Bertucci
 Guido Cagnacci - the museum recently acquired (2005) his work "Allegoria dell'Astrologia sferica"
 Antonio Canova
 Baldassarre Carrari
 Giovanni Crivelli
 Domenichino
 Giovanni Fattori
 Guercino
 Lorenzo di Credi
 Carlo Magini
 Girolamo Marchesi
 Melozzo da Forlì(?)
 Francesco Menzocchi
 Livio Modigliani
 Giorgio Morandi
 Marco Palmezzano
 Adolfo Wildt

It contains the Verzocchi collection of 20th-century Italian painting.

Note

External links

Fuller account
Works exhibited

Art museums and galleries in Emilia-Romagna
Buildings and structures in Forlì
Museums in Emilia-Romagna

redpajama_set_name: RedPajamaWikipedia


text:
Listen: Phoenix, Mike Groom, Vegas discussed on Mike Broomhead
"On our news, 92 3 FM and the news Thanks so much for being here it is. It's Mike Groom had show way. Appreciate you being with us homos problem in Arizona is growing and it we have grown in the city of Phoenix in homelessness in a while, and there are a lot of groups to do a great job. You've heard me talk about them before. Because other major cities it's seen a lot more in Phoenix. It's kind of in an area of downtown In 500. People sleep every night in sleeping bags, tents, makeshift tarp shelters on the streets around human services campus in south of downtown Phoenix. And that's cast. That's this campus. That's downtown. But what covert 19 has done is this homeless People used to be able to go into a respite centre during the day and get out of the heat. They can't because of covert 19. They had to stay outside in the shade. Sleeping conditions used to be. They were packed together well because of covert 19. They put up chains and they made it harder so that people would distance and so the homeless camps are now spread out in a bigger area. People are distanced more And the reason why I'm bringing this up is specifically about covert 19, but about the problem of homelessness that is about to come to us. And I think we're going to start seeing more and more people, unexpected people becoming homeless, at least for a short time because of what covert 19 has done to our economy. We have put a moratorium on evictions. But that's not going to last forever. And although some people be able to cut deals with landlords and get their jobs back and climb out of that hole, we understand statistically, there will be people. That right now have homes that will not in the future, And it's going to make this problem even worse. So the story that I get this from is from a part of downtown where the business is say, you know, we've been We have been apart of this forever and really, the city isn't doing a whole lot about it. I don't think that it's a municipal problem. Um And I don't think, um, I think it's done by the private sector in a better way. I've been down to cast I I've mentioned before I friend Marty. It circled the city and what they're doing The great work they do, Andi just it's it's all privately funded, I would say, mostly privately funded. And to see the different holes that they fill when it comes to homeless people is remarkable and they deserve our attention and our support. I've told this story before that I had a cousin who was my hero. When we were little kids. He was the older brother. I didn't have. We spent summers together and and we grew up together until I moved to Florida when I was very young when we stayed in touch. When his mother died When he was in high school, he went down the road of drugs and alcohol, and he drowned his sorrows that way and became homeless lived in Vegas, where my his oldest sister, My oldest cousin lived, and when I would go in visit, I would always find him. He always knew I was coming and we would hang out. And he looked like a typical homeless person, but not not to me. He was just my cousin. And That for me became the face of homelessness. My cousin died in a homeless camp in Vegas. Whether it was an overdose or a seizure. He died. And so when I see the face of a homeless person, I know there's a story and there are people that love them, and now my cousin was homeless by his own devices. He caused the problem. He didn't want to fix the problem. He wanted to stay on the streets. But there are many people out there that want and need the help that we can give them. We're going to see families on the street. We're going to see more and more people looking for shelter and they're not going to find it. This is not a problem. It's going to go. That's going to go away without the help of people like you and me. And if you think it doesn't affect you, it does And it will. And until someone in your life touches you that's homeless until someone you know is without a home. You may be human. It may not be a problem that moves you. But there are people out there in the private sector that are working very hard to solve parts of these problems. Then there's ways we can help the three things we offer. The world is money, talent in time, and those three things we have the different degrees at different times in our life. And if you're involved in a cause that you believe in, I think that's awesome. If you're looking for girl Group of people to get involved with these organisations, whether it's ST Vincent de Paul Circle the city they're giving masks to homeless people. When homeless people are Covad 19 positive. They're testing them. They're putting them places that are separated, whether their pods or sometimes these old hotels where they're giving them rooms so that the Koven 19 problem. Would spread like wildfire throughout the homeless community. These organizations are making sure that doesn't happen. They're getting some basic medical care. They give them hygiene kits so they can have the basics of toothbrushes and toothpaste and a mask. And there's a lot of work that's happening to make sure that society isn't spreading covert 19 through the homeless community and at the same time trying to fill their daily needs of just nourishment in water. And maybe cure the problem for those that are looking, and many of these groups are doing a great job off, making sure that they can get off the streets. Whatever their issues are, whether it's a money or a legal problem that keeps him from getting a driver's license, a transportation issue a drug issue or a mental health issue, helping these people overcome these basic problems and then getting them back. Into a home and back into the mainstream of society. That's something that I want to be a part of, But I bring it up because in the coming months when covert 19 is over in the moratorium on evictions end we're going to see more and more people face homelessness. And it's going to be people just like you and me. And we could help stave that problem off right now. If we get involved find an organization that's fighting it and get involved. Nancy Pelosi unloads on the media coming up after the next hour. We're going to talk about her unloading and what she said in wide. Trading wealth is in place for you and your retirement, and they got advice for you. If you've been separated from a job for whatever reason during covert 19, and you still have money with a former employer in there for a one care retirement account, get control of those dollars. You want to have control of those dollars because if they close or sell your money could be tied up for a long time without your control. And then trading wealth is going to help you stay on track for retirement. What's great about this? Is there going to help you make sure that you're navigating the good times. You're prepared for the times that we were unexpected, and when it's time for you to retire, you have."
KTAR 92.3FM
Phoenix, Mike Groom, Vegas discussed on Mike Broomhead
Mike Broomhead
1 year ago | KTAR 92.3FM
Phoenix Mike Groom Vegas Arizona Nancy Pelosi Covad St Vincent De Paul Circle Marty Florida
On our news, 92 3 FM and the news Thanks so much for being here it is. It's Mike Groom had show way. Appreciate you being with us homos problem in Arizona is growing and it we have grown in the city of Phoenix in homelessness in a while, and there are a lot of groups to do a great job. You've heard me talk about them before. Because other major cities it's seen a lot more in Phoenix. It's kind of in an area of downtown In 500. People sleep every night in sleeping bags, tents, makeshift tarp shelters on the streets around human services campus in south of downtown Phoenix. And that's cast. That's this campus. That's downtown. But what covert 19 has done is this homeless People used to be able to go into a respite centre during the day and get out of the heat. They can't because of covert 19. They had to stay outside in the shade. Sleeping conditions used to be. They were packed together well because of covert 19. They put up chains and they made it harder so that people would distance and so the homeless camps are now spread out in a bigger area. People are distanced more And the reason why I'm bringing this up is specifically about covert 19, but about the problem of homelessness that is about to come to us. And I think we're going to start seeing more and more people, unexpected people becoming homeless, at least for a short time because of what covert 19 has done to our economy. We have put a moratorium on evictions. But that's not going to last forever. And although some people be able to cut deals with landlords and get their jobs back and climb out of that hole, we understand statistically, there will be people. That right now have homes that will not in the future, And it's going to make this problem even worse. So the story that I get this from is from a part of downtown where the business is say, you know, we've been We have been apart of this forever and really, the city isn't doing a whole lot about it. I don't think that it's a municipal problem. Um And I don't think, um, I think it's done by the private sector in a better way. I've been down to cast I I've mentioned before I friend Marty. It circled the city and what they're doing The great work they do, Andi just it's it's all privately funded, I would say, mostly privately funded. And to see the different holes that they fill when it comes to homeless people is remarkable and they deserve our attention and our support. I've told this story before that I had a cousin who was my hero. When we were little kids. He was the older brother. I didn't have. We spent summers together and and we grew up together until I moved to Florida when I was very young when we stayed in touch. When his mother died When he was in high school, he went down the road of drugs and alcohol, and he drowned his sorrows that way and became homeless lived in Vegas, where my his oldest sister, My oldest cousin lived, and when I would go in visit, I would always find him. He always knew I was coming and we would hang out. And he looked like a typical homeless person, but not not to me. He was just my cousin. And That for me became the face of homelessness. My cousin died in a homeless camp in Vegas. Whether it was an overdose or a seizure. He died. And so when I see the face of a homeless person, I know there's a story and there are people that love them, and now my cousin was homeless by his own devices. He caused the problem. He didn't want to fix the problem. He wanted to stay on the streets. But there are many people out there that want and need the help that we can give them. We're going to see families on the street. We're going to see more and more people looking for shelter and they're not going to find it. This is not a problem. It's going to go. That's going to go away without the help of people like you and me. And if you think it doesn't affect you, it does And it will. And until someone in your life touches you that's homeless until someone you know is without a home. You may be human. It may not be a problem that moves you. But there are people out there in the private sector that are working very hard to solve parts of these problems. Then there's ways we can help the three things we offer. The world is money, talent in time, and those three things we have the different degrees at different times in our life. And if you're involved in a cause that you believe in, I think that's awesome. If you're looking for girl Group of people to get involved with these organisations, whether it's ST Vincent de Paul Circle the city they're giving masks to homeless people. When homeless people are Covad 19 positive. They're testing them. They're putting them places that are separated, whether their pods or sometimes these old hotels where they're giving them rooms so that the Koven 19 problem. Would spread like wildfire throughout the homeless community. These organizations are making sure that doesn't happen. They're getting some basic medical care. They give them hygiene kits so they can have the basics of toothbrushes and toothpaste and a mask. And there's a lot of work that's happening to make sure that society isn't spreading covert 19 through the homeless community and at the same time trying to fill their daily needs of just nourishment in water. And maybe cure the problem for those that are looking, and many of these groups are doing a great job off, making sure that they can get off the streets. Whatever their issues are, whether it's a money or a legal problem that keeps him from getting a driver's license, a transportation issue a drug issue or a mental health issue, helping these people overcome these basic problems and then getting them back. Into a home and back into the mainstream of society. That's something that I want to be a part of, But I bring it up because in the coming months when covert 19 is over in the moratorium on evictions end we're going to see more and more people face homelessness. And it's going to be people just like you and me. And we could help stave that problem off right now. If we get involved find an organization that's fighting it and get involved. Nancy Pelosi unloads on the media coming up after the next hour. We're going to talk about her unloading and what she said in wide. Trading wealth is in place for you and your retirement, and they got advice for you. If you've been separated from a job for whatever reason during covert 19, and you still have money with a former employer in there for a one care retirement account, get control of those dollars. You want to have control of those dollars because if they close or sell your money could be tied up for a long time without your control. And then trading wealth is going to help you stay on track for retirement. What's great about this? Is there going to help you make sure that you're navigating the good times. You're prepared for the times that we were unexpected, and when it's time for you to retire, you have.

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: Alternative to 'ban' for Varnish 2.x We are using below code in Varnish 4.x:
 if (req.http.X-Pool) {
    ban("obj.http.X-Pool ~ " + req.http.X-Pool);
 }

Now we are moving to Fastly which uses Varnish 2.x, so we are not getting what is the alternative to ban in Varnish 2.x

A: For help with Fastly Varnish/VCL I recommend reading through both:

*

*https://developer.fastly.com/

*https://www.integralist.co.uk/posts/fastly-varnish/
I would also generally recommend reaching out to support@fastly.com (they're a good group of people).
With regards to your question, I'm unfamiliar with bans in standard Varnish, but reading through https://varnish-cache.org/docs/trunk/users-guide/purging.html#bans it suggests that a ban is a way to prevent cached content from being served.
So the solution depends on what you're trying to achieve when that happens.
If you just want to avoid the cache you can return a pass from the various subroutines like vcl_recv and vcl_hit (although from vcl_hit will cause a hit-for-pass and so that will cause the request to go straight to the backend.
You could also add custom logic to either vcl_recv or maybe even vcl_hit  (if you want to be sure the content requested was actually cached) and from there you could trigger an error which would send you to vcl_error where you could construct a synthetic response:
sub vcl_hit {
  #FASTLY hit

  if (<some_condition>) {
    error 700
  }
}
sub vcl_error {
  #FASTLY error

  if (obj.status == 700) {
    set obj.status = 404;

    synthetic {"
      <!doctype html>

          <meta charset="utf-8">
          <title>Error</title>

404 Not Found (varnish)

      "};

    return(deliver);
  }

Alternatively from either vcl_recv or vcl_hit you might want to restart and then have a check for the restart and do something different (change the backend or the request in some way):
sub vcl_recv {
  #FASTLY recv

  if (req.restarts > 0) {
    if (<some_condition>) {
      // do something
    }
  }
}

redpajama_set_name: RedPajamaStackExchange


text:
Q: Push update in real time to a bower via process? I have a script written in PHP and it run in the background on Linux.
Whatever happening in the script, it insert a log into a database.
I could use ajax to refresh every 3 seconds on the browser to pull update via database.
However, I prefer a real time push a browser, how can that be done?


Centre Marketg was built in 1853/54 but the tower was declared unsafe in the early 1860's and removed. The tower served as the city's first fire alarm bell, which wighed 6,000 pounds. Watchmen were stationed in it to strike the alarm in case of a fire. When the tower was declared unsafe, the bell was relocated at Market, Arlington, Augusta & Nicholson Streets. The second floor of Centre Market held the city offices. The Common Council Chamber was brilliantly lighted by gas from two large chandeliers. The building and surrounding land cost a little less than $90,000.


Since the Cha-Cha silhouette is so popular, we've re-imagined it to create our most shiny and shimmery version yet! The two-tiered skirt allows for plenty of twirling opportunities, while the bodice features gold lamé jacquard in a metallic leaf pattern that's a perfect fit for the season. Additional features include a soft V-neck and zipper closure at the back for fuss-free dressing.


I was pretty new to comics when I got burned on some not-so-great books about characters I loved. I told myself to just follow reviews at that point and started combing the internet for iconic runs. Green Lantern was never on my radar as an interesting character. The premise of "make anything with this ring" seemed silly. It didn't take many searches for me to realize that Geoff Johns' Blackest Night run was unmissiable, however, and I caved and decided to dive in with both feet. I started picking up the trades as cheap as I could and didn't start reading the series until I had a good 4-5 books. When I did, I fully understood the near-consensual praise for the run. I still don't love the character but Johns' wrote the book so well that I cared and was invested in him. Genius concepts, deep plotting and great pacing all ended with the great event book "Blackest Night". This was aided by amazing art by talents like Ivan Reis and Ethan Van Sciver that gave depth to the entire Green Lantern universe. John's run continued after the event, but I found that it made for the perfect climax for his overall arc.


This beautifully remodeled Naples Island home has it all. Sitting on a 4,568 square foot lot you have wonderful indoor to outdoor living. This home is loaded with high quality details. The dining area is open to the modern kitchen complete with high quality stainless steel appliances including 5-burner stove, granite counter tops, tiled backsplash and custom wood cabinetry that offers ample storage space. The original hardwood floors have been vibrantly restored laying the groundwork to showcase everything from new baseboards, custom doors, recessed lighting and ceiling fans. Previous renovation included new electrical, plumbing and roof. Exquisite contemporary bathroom spared no expense with double vanity, Robern uplifting mirrors, vanity wall sconces, custom built cabinetry and stunning tiled shower. Oversized double pane windows bring in natural light and ocean breezes. French Doors lead out to a spacious drought friendly low maintenance backyard retreat. Abundant parking options including 2 car garage with electric car charging station, additional parking for up to 3 cars, a boat or other toys. Excellent Location!! Walk to Fine dining and shops on 2nd St., Mother's Beach, Marine Stadium, Naples Canals, Alamitos Bay and more!


A new piece just added to the site here. You customize with your word that follows the i am. What do you think, you like? What would your i am be?
This is absolutely gorgeous! I'm seriously going to have to look into owning one!


About to take-off, fold back from your extension and pull up the bars in front of you to lift the bike up while tucking your knees up to clear the way for the bike. Pointing your feet down, apply back-force on the pedals and level up the bike horizontally to lift the rear wheel further up.
1° Build up enough momentum to clear the obstacle length during the duration of your bunny hop, then stop pedaling and lean forward.
4° Carry on with a full jump impulse, keeping your arms straight to maintain pressure on the ground while moving up on the high pedal position.
6° Lift the rear wheel further to level up the bike by gripping up the pedals with your soles (applying back-force), and tuck until you have cleared the obstacle.
On the bike, just do the same, crouch to flex before performing a firm extension on the pedals as they move up (the bike stands up as you push on your legs and pull the bars to your abdomen, which also boosts your own jump).
Push on the ground before take-off.
During your extension, keep your arms stretched (eventually they'll end up along your torso) so you keep pushing firmly on the ground (through the rear wheel) for as long as possible. When you have reached the full extension, tuck your knees while pulling on the handlebars to lift the whole bike up in front of you.
In a perfectly straight bunny hop, you may find that as you level the bike up, the rear tyre hits you in the bum and limits the overall bike lift (and bunny hop height). By tucking the bike slightly sideways to your front foot side, you can clear your bum out of the way to your back-foot side, letting the rear wheel move further up.
That little swing sideways will not only gain you a few extra inches, it will also help you clear the obstacle sooner.


It's pretty apparent that Superhero films are the wave right now, and many directors are trying to get a piece of the action by picking up heroes to bring to the big screen, and that includes The Neon Demon and Drive Director Nicolas Refn.
So a Batgirl movie. It's definitely not outside the realms of possibility. We've already seen Batman in the DCEU, and we know he's been around for a few years. On top of that, J.K. Simmons has been cast as Commissioner Gordon, who is the father of Barbara Gordon aka Batgirl, so introducting her character wouldn't be too hard. Obviously this film would have to come out some time after Justice League Part One in 2017, but we at least have a director who'd be willing to bring her character to life. Now it's just your move, Warner Bros..
Previous RUMOR PATROL: Marvel & Fox In Early Talks For A Crossover?


A roughly taxonomised listing of some of the papers on Homotopy Type Theory. Titles link to more details, bibdata, etc. Currently very incomplete; please add!
Homotopy-Theoretic Models of Type Theory. Peter Arndt and Chris Kapulkin. In Typed Lambda Calculi and Applications, volume 6690 of Lecture Notes in Computer Science, pages 45?60.
A coherence theorem for Martin-Löf?s type theory. Michael Hedberg, Journal of Functional Programming 8 (4): 413?436, July 1998.
Revision on March 28, 2015 at 13:01:53 by Mike Shulman. See the history of this page for a list of all contributions to it.


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text:
About Saint John & Surrounding Areas
The City of Saint John, New Brunswick was built on the banks of the St. John River and Saint John Harbour, on top of some of the oldest geology in the world, Saint John, the first incorporated city in Canada (1785), is the largest city on the Bay of Fundy and the industrial backbone of New Brunswick.
For centuries, our harbour has been a haven for people from across the globe. For many of these people, this place, born of the water, became home. Today 67,000 residents still enjoy easy access to the Bay, the St. John River and the lakes around the city.
• A city of original neighbourhoods, families and individuals who live in Saint John can choose from a variety of housing.
• Through school districts Anglophone South School District and District scolaire francophone Sud, children and youth enjoy quality public education.
• UNB Saint John and New Brunswick Community College (NBCC) Saint John offer a variety of post secondary education programs.
• The City's Uptown is a hub for arts, culture and heritage, and our recreational and cultural facilities are among the best in the Atlantic provinces.
• Saint John's East side shopping district is the largest single concentration of shopping anywhere in the Maritimes. The City's West side is also undergoing retail expansion.
• Saint John is home to the province's heart and trauma centre and the largest hospital in New Brunswick.
• Saint John has the largest and most used transit system in New Brunswick.
(http://www.saintjohn.ca/en/home/About/oursj/default.aspx)
Saint John, New Brunswick
Properties For Sale In and Around Saint John
Showing 121 to 180 of 889 properties
Listings last updated Wednesday July 17 2019
ID#: NB022788
182 Germain Street, Saint John, New Brunswick (id Nb022788)
Courtesy of KELLER WILLIAMS CAPITAL REALTY (S223)
24 Big K Ranch Lane Lot 16-13, Queenstown, New Brunswick (id Nb018549)
Courtesy of RE/MAX East Coast Elite Realty
- Drury Cove Road, Saint John, New Brunswick (id Nb019558)
Courtesy of COLDWELL BANKER SELECT REALTY (S214)
157 Bay Crescent, Saint John, New Brunswick (id Nb023677)
Courtesy of RE/MAX PROFESSIONALS SAINT JOHN INC. (S119)
- Paradox Point, Bocabec, New Brunswick (id Nb015651)
Courtesy of FUNDY BAY REAL ESTATE GROUP INC. (S188)
48 Waterford Lane, Fredericton, New Brunswick (id Nb022232)
Courtesy of Keller Williams Capital Realty
19 Oleary Crescent, Saint John, New Brunswick (id Nb019617)
61 Crawford Road, Saint John, New Brunswick (id Nb026110)
35 Barrett Court, Fredericton, New Brunswick (id Nb023140)
1595 Manawagonish Road, Saint John, New Brunswick (id Nb021144)
Courtesy of EXIT REALTY SPECIALISTS ROTHESAY (S201)
- Ash Street, Saint John, New Brunswick (id Nb013330)
Courtesy of ROYAL LEPAGE ATLANTIC (S105)
3115 Route 845, Long Reach, New Brunswick (id Nb023240)
12 Rowan Berry Terrace, Saint John, New Brunswick (id Nb027545)
145 Boars Head Road, Saint John, New Brunswick (id Nb027627)
7 Blanchard Lane, Saint John, New Brunswick (id Nb018957)
8 Goddards Lane, Grand Lake, New Brunswick (id Nb008466)
169 Manners Sutton Road, Saint John, New Brunswick (id Nb028006)
Courtesy of EXIT REALTY SPECIALISTS (S181)
5 Westmount Drive, Saint John, New Brunswick (id Nb026111)
590 Queen Street Unit# 607, Fredericton, New Brunswick (id Nb025975)
13716 105 Route, Tilley, New Brunswick (id Nb023683)
Courtesy of CENTURY 21 ALL SEASONS REALTY
10-12 City Line, Saint John, New Brunswick (id Nb008227)
11 Thomas Road, Saint John, New Brunswick (id Nb019831)
- Norris Road, Saint John, New Brunswick (id Nb017067)
49 By Road #5, Saint John, New Brunswick (id Nb019596)
- Woolastook Drive, Grand Bay-westfield, New Brunswick (id Nb013785)
243 Prince William Street Unit# 302, Saint John, New Brunswick (id Nb019591)
3424 Route 105, Bear Island, New Brunswick (id Nb011190)
Courtesy of Gardiner Realty Royal LePage Fredericton.
Courtesy of CENTURY 21, A&T COUNTRYSIDE REALTY INC. (110)
26 Meadow Drive, Darlings Island, New Brunswick (id Nb027696)
15 Main Street West, Saint John, New Brunswick (id Nb023990)
19 Marlin Drive, Saint John, New Brunswick (id Nb022836)
248 Anthony's Cove Road, Saint John, New Brunswick (id Nb008169)
ID#: SJ174882
- Route 850, Kingston, New Brunswick (id Sj174882)
2156 Route 124, Hatfield Point, New Brunswick (id Nb018573)
Courtesy of RE/MAX QUALITY REAL ESTATE (S208)
7878 Route 112, Hunters Home, New Brunswick (id Nb023160)
12 Cavanagh Court, Saint John, New Brunswick (id Nb023193)
47-49 Cranston Avenue, Saint John, New Brunswick (id Nb025844)
436 Gondola Point Road, Quispamsis, New Brunswick (id Nb021396)
72 Dantes Drive, Saint John, New Brunswick (id Nb025477)
309 Hawthorne Avenue, Saint John, New Brunswick (id Nb026512)
309 Hawthorne Ave Ext, Saint John, New Brunswick (id Nb026513)
9 Vista Ridge Crescent, Saint John, New Brunswick (id Nb022607)
2971 Route 790, Musquash, New Brunswick (id Nb022488)
85-87 Clarendon Street, Saint John, New Brunswick (id Nb018225)
34 Noel Avenue, Saint John, New Brunswick (id Nb023248)
48 Gaelic Drive, Saint John, New Brunswick (id Nb022397)
44 Northumberland Avenue, Saint John, New Brunswick (id Nb019336)
593 Latimore Lake Road, Saint John, New Brunswick (id Nb013619)
696-700 Rothesay Avenue, Saint John, New Brunswick (id Nb001632)
20 O'leary, Saint John, New Brunswick (id Nb023635)
35 Edgemount, Grand Bay-westfield, New Brunswick (id Nb025593)
19 Third Street, Saint John, New Brunswick (id Nb025891)
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The most common method by which radon enters a residential property is through the air, sucked in from the soil through tiny holes and cracks in the building's foundations or basement walls. However, that is not the only way that this potentially harmful, radioactive gas can infiltrate your home. Another common method by which radon enters residential properties is through the water supply. Waterborne radon can be just as harmful to health as air-borne radon. And because it is tasteless, colorless, and odorless, you will not know whether radon is present in your water supply unless you test for it.
We understand how worrying it can be to think that your health and the health of your family may be at risk from a radioactive gas that you cannot see, smell, or taste. Our highly-qualified contractors are trained to test for waterborne radon contamination. We can visit your WV home and produce a report on waterborne radon that will enable you to decide on your next course of action.
If high levels of waterborne radon are identified, we can provide advice on implementing an effective radon mitigation program to prevent further radon contamination and to reduce existing levels of radon.
Throughout West Virginia, the soil is layered with naturally-occurring rocks which produce radon as they break down. The radon permeates the soil and is emitted as a gas which disperses into the atmosphere, unless it becomes trapped in a structure. In this way, radon gas can enter your home through vents and small cracks in the foundations and basement walls.
However, the radon in the soil can also contaminate the ground water. This ground water feeds into wells, which may provide domestic water to houses in the Beckley area. If you draw your household water from a well, it is highly-recommended that you have it tested for radon to protect your health and that of your family. Contaminated ground water can also allow radon to seep in to your water pipes that feed your washing machine and dishwasher.
While it is worrying to think that you may be exposing yourself and your family to radon every time you turn a tap or wash the dishes, there are steps that you can take.
By contacting S.W.A.T. to arrange to a waterborne radon test, you can be sure of the extent of the problem. You can then put in place an appropriate radon mitigation plan to reduce current levels of radon and prevent future contamination.
The state of West Virginia requires radon mitigation contractors to hold a valid contractor's license prior to carrying out radon mitigation work. In addition, they must have completed trade-related examinations or possess additional qualifications related to radon mitigation. All of our employees are fully trained and certified to conduct waterborne radon mitigation.


Long ago when I was a law professor, I was at a conference and a man I didn't recognize greeted me warmly. He said he wanted to thank me for changing his life.
I was embarrassed as I listened to him tell me that he had met me after a speech I had given at his law school. He said he was discouraged and disheartened about ever becoming a lawyer and that he was ready to quit. But I had counseled him and he had decided to stick it out.
He said he had been looking for me so he could tell me personally that not only had he graduated but had just become the nation's first Mexican-American law school dean.
His decision to seek me out to share the story was a generous and much-appreciated gift I'll never forget. But his description did not ignite my memory and I felt shallow and ashamed. Here this fellow was expressing deep gratitude for something I could not recall. I only began to forgive myself when I realized I didn't remember the incident because I engaged in this sort of discussion with students often, and this conversation was not extraordinary for me. That's what teachers are supposed to do: share their knowledge, provide alternative perspectives and offer encouragement and inspiration.
"Teachable moments" often arise unannounced and unnoticed. Teachers and parents hardly ever know what will be remembered and what will be forgotten. But if we want to have an occasional lasting impact we have to have faith that at least some of the things we say will really matter. And since we can't always know what those things are, we have to presume that everything we say will matter.
So true, Michael; it reminds me that all too often what I say just to myself is not as healthy, ethical, and honorable as it could be. Thanks partly to you, I try to better at all my thoughts and words, whether they're heard in forest/marketplace or not. Thanks.
Thanks Abe for being on the journey with me. We're both getting better.


It puts the aspirations and needs of those who produce, distribute and consume food at the heart of food systems and policies rather than the demands of markets and corporations. It defends the interests and inclusion of the next generation. It offers a strategy to resist and dismantle the current corporate trade and food regime, and directions for food, farming, pastoral and fisheries systems determined by local producers and users.
Food sovereignty prioritises local and national economies and markets and empowers peasant and family farmer-driven agriculture, artisanal – fishing, pastoralist-led grazing, and food production, distribution and consumption based on environmental, social and economic sustainability.
Food sovereignty promotes transparent trade that guarantees just incomes to all peoples as well as the rights of consumers to control their food and nutrition. It ensures that the rights to use and manage lands, territories, waters, seeds, livestock and biodiversity are in the hands of those of us who produce food.
Nyeleni Europe – the widest international movement aiming to realize food sovereignty in Europe. Boerengroep is part of the Dutch Delegation of Nyeleni Europe Forum for Food Sovereignty that will take place in October 2016, Romania.
Via Campesina – the largest international peasant movement, which launched the concept of Food Sovereignty in 1996 during the FAO World Food Summit in Rome. La Via Campesina is an international movement bringing together millions of peasants, smalland medium size farmers, landless people, rural women and youth, indigenous people, migrants and agricultural workers from around the world. Built on a strong sense of unity, solidarity between these groups, it defends peasant agriculture for food sovereignty as a way to promote social justice and dignity and strongly opposes corporate driven agriculture that destroys social relations and nature. See also their flyer.
List on Food Sovereignty here..
Food Sovereignty: Towards democracy in localized food systems. FIAN International. Here..
Food Sovereignty comes of Age. Africa leads efforts to rethink our food system. by UK Food Goup. Download here..
Developing and strengthening the concept of peoples' food sovereignty. Read here..
Peoples' Food Sovereignty Statement. Read here..
Food Sovereignty Framework: Concept and Historical Context. Read here..
Via Campesina, 2003. Food Sovereignty: Global Rallying Cry of Farmer Movements. Read here..
GRAIN, 2005. Food Sovereignty: turning the global food system upside down. Read Here..
Why a convention on food sovereignty. Read here..
Put food sovereignty at the core of the EU CAP. Brussels, 2010. Download here..
NYÉLÉNI 2007. 27 February 2007 Declaration of the Forum for Food Sovereignty, Nyéléni, Sélingué, Mali.
the article by Via Campesina Over 15 years' struggle for recognition of peasant rights in the international human rights system.
Cetim.ch has been committed for many years to better protection and promotion of peasant rights, alongside La Via Campesina and FIAN. They have published a series of documents on the right to land, economic, social and cultural rights and civil and political rights.
VIDEO Hands on the land for food sovereignty and climate justice. Frontline communities and social movements around the world explain why we need to keep our hands on the land for food sovereignty and climate justice! They talk about the false solutions they are presented with against climate change, and the real solutions and ways forward that small-scale food producers promote.
VIDEO What does it mean to be peasant in Romania? With half of the EU's peasants, Romania is the ideal place for the largest pan-European food sovereignty forum in history. Hundreds of farmers, peasants, pastoralists, fishers, rural youth, environmentalists and others gathered in Cluj-Napoca between the 25th and 30th of October 2016 to build common strategies in order to re-organise the way we structure our society around food and agriculture.